Chemistry study guide

Chemistry Concentration Basics

Q: When can I use the dilution equation?

Use c_1V_1=c_2V_2 when the same amount of the specified solute is conserved and the two concentrations are expressed on compatible volume-based bases. It does not account for reaction, precipitation, loss of solute or an added amount of solute. Mass percent and molality have different denominators, so their numerical values cannot simply replace molarity in that equation. A target above the stock requires concentration rather than adding solvent.

Q: Is neutral pH always seven, and do displayed digits show uncertainty?

In the ideal water model, neutral pH is mathrm pK_mathrm w/2. It is 7.00 when the adopted water constant has mathrm pK_mathrm w=14.00, approximately the common aqueous reference at 25 ^°mathrm C. Different conditions can change that constant. The calculator's decimal places control display, not measurement uncertainty. Keep the temperature, model, original readings and reporting convention visible when interpreting an answer.

Connect mass, moles and solution volume through one worked example. Then check dilution, serial dilution, concentration labels and pH with clear units and visible assumptions.

Verified nominal calculations · Three original illustrations · Prefilled calculator examples

Try the dilution example

Name the solute and the denominator first

Concentration describes the composition of a solution, but the number alone is incomplete. State which component is being counted and whether the denominator is solution volume, solution mass or solvent mass. The same sample can have different numerical concentrations on different bases without any contradiction.

c=\frac{n_{\mathrm{solute}}}{V_{\mathrm{solution}}}

Here $c$ is amount concentration, commonly called molarity; $n$ is amount of the specified solute; and $V_{\mathrm{solution}}$ is the final solution volume. We use $M_{\mathrm m}$ for molar mass, so its symbol cannot be confused with the common molarity unit $\mathrm M$ . The denominator is not simply the volume of water added. OpenStax introduces this solution-volume definition.

The mole is the SI unit of amount of substance. Its entities must be specified: atoms, molecules, ions or formula units are not interchangeable descriptions. For the sodium chloride example below, $\mathrm{NaCl}$ identifies the analytical solute formula; it does not mean dissolved salt consists of intact sodium chloride molecules. The BIPM mole definition makes that entity requirement explicit.

Concentration workflow identifying the solute and basis, converting mass and volume, conserving solute in dilution and stating the pH model. — Start with a specified solute and denominator, then choose a compatible concentration relationship and state the assumptions.

Convert molarity prefixes before dividing

Prefixes change the numerical scale, not the concentration basis. A millimolar number is a thousand times the numerical molar value for the same solution. A micromolar number is a million times the molar value. Keep the prefix attached to the unit throughout a conversion.

On small screens, swipe the reference tables or focus them and use the arrow keys. Wide equations also support keyboard and touch scrolling.

Different prefixes, the same amount-per-solution-volume basis
Name	Symbol	Definition
Molar	$\mathrm M$	$1\,\mathrm M=1\,\mathrm{mol/L}$
Millimolar	$\mathrm{mM}$	$1\,\mathrm{mM}=10^{-3}\,\mathrm{mol/L}$
Micromolar	$\mathrm{\mu M}$	$1\,\mathrm{\mu M}=10^{-6}\,\mathrm{mol/L}$
Nanomolar	$\mathrm{nM}$	$1\,\mathrm{nM}=10^{-9}\,\mathrm{mol/L}$

0.4000\,\mathrm{mol/L}=400.0\,\mathrm{mM}

25.0\,\mathrm{mM}=0.0250\,\mathrm{mol/L}

The prefix conversions are exact definitions. They do not make a measured concentration exact, and they do not change the intended significant figures. Dividing the bare numbers $0.4000$ and $25.0$ before reconciling their units would lose a factor of $1000$ . The Volume Converter can separately check the solution-volume conversion.

Worked example: convert a mass to molarity

These are hypothetical teaching inputs. Assume pure anhydrous sodium chloride and a stated final solution volume. A hydrate, a different chemical formula or an assay below full purity would require a different mass-to-amount calculation. No composition or purity correction is inferred.

1. Adopt and identify the molar mass

M_{\mathrm m}(\mathrm{NaCl})=58.44\,\mathrm{g/mol}

The calculator uses nominal sodium and chlorine values $22.990$ and $35.45$ from the CIAAW abridged table, giving the adopted sum $58.44\,\mathrm{g/mol}$ . This is a working molar mass, not an exact physical constant or a propagated atomic-weight uncertainty. Check the CIAAW reference values and inspect the formula and mass-to-moles calculation.

2. Convert the sample mass to amount

\begin{aligned}n&=\frac{m}{M_{\mathrm m}}\\&=\frac{5.844\,\mathrm g}{58.44\,\mathrm{g/mol}}\\&=0.1000\,\mathrm{mol}\end{aligned}

Grams cancel against grams per mole, leaving moles. The nominal decimal-input arithmetic is checked without rounding intermediates; the reported value here keeps four significant figures under the introductory reporting convention.

3. Use the final solution volume in liters

V_{\mathrm{solution}}=250.0\,\mathrm{mL}=0.2500\,\mathrm L

c=\frac{0.1000\,\mathrm{mol}}{0.2500\,\mathrm L}=0.4000\,\mathrm{mol/L}

The specified final volume includes the dissolved solute. Describing the same amount as added to that volume of solvent would be a different problem. Temperature can also change a solution's volume, so record the conditions when a concentration is being measured rather than merely calculated from a worksheet.

4. Cross-check the mass concentration

c_{\mathrm{mass}}=\frac{5.844\,\mathrm g}{0.2500\,\mathrm L}=23.376\,\mathrm{g/L}

That last value retains guard digits for the cross-check; a four-significant-figure report is $23.38\,\mathrm{g/L}$ . Converting the unrounded mass concentration back through the same adopted molar mass gives $0.4000\,\mathrm{mol/L}$ . Check the mass-to-molar unit bridge. This prefilled conversion keeps the same initial and final volume, so no dilution is applied.

Worked example: dilute a stock aliquot

Use the $0.4000\,\mathrm{mol/L}$ stock above to calculate the aliquot needed for $100.0\,\mathrm{mL}$ of a $25.0\,\mathrm{mM}$ solution. Assume the same solute is conserved: no reaction, precipitation, solute loss or additional solute. Only an aliquot of the prepared stock enters this calculation.

A sixteenfold dilution of a stock aliquot

The top scale measures volume; the bottom scale measures concentration. Each starts at zero. The $6.25\,\mathrm{mL}$ stock aliquot and $100.0\,\mathrm{mL}$ diluted solution each contain $2.50\,\mathrm{mmol}$ in the nominal model. This is the aliquot's solute, not the amount in the entire prepared stock.

1. Normalize the units

c_1=0.4000\,\mathrm{mol/L}

c_2=0.0250\,\mathrm{mol/L}

V_2=0.1000\,\mathrm L

2. Conserve the amount in the aliquot

c_1V_1=c_2V_2

Concentration times solution volume gives solute amount on each side. This relationship compares compatible concentration bases; the symbols are not permission to substitute a percentage with a different denominator.

3. Solve for the stock volume

\begin{aligned}V_1&=\frac{c_2V_2}{c_1}\\&=\frac{(0.0250\,\mathrm{mol/L})(0.1000\,\mathrm L)}{0.4000\,\mathrm{mol/L}}\\&=0.00625\,\mathrm L=6.25\,\mathrm{mL}\end{aligned}

The nominal dilution factor is $D=c_1/c_2=V_2/V_1=16$ . The lower target concentration therefore requires a final volume sixteen times the aliquot volume. The computed $6.25\,\mathrm{mL}$ is the stock aliquot; $100.0\,\mathrm{mL}$ is the final solution volume. Open this stock-volume calculation.

4. Check the conserved amount

\begin{aligned}c_1V_1&=(0.4000\times0.00625)\,\mathrm{mol}\\&=0.00250\,\mathrm{mol}\end{aligned}

\begin{aligned}c_2V_2&=(0.0250\times0.1000)\,\mathrm{mol}\\&=0.00250\,\mathrm{mol}\end{aligned}

Both states contain $2.50\,\mathrm{mmol}$ in the nominal aliquot calculation. This does not transfer the entire stock's $0.1000\,\mathrm{mol}$ . An unrounded nominal volume difference is $(100.0-6.25)\,\mathrm{mL}=93.75\,\mathrm{mL}$ ; the classroom subtraction rule would report that difference as $93.8\,\mathrm{mL}$ . Neither figure guarantees an independently measured solvent addition because liquid volumes need not be additive.

The same example can solve four different unknowns

Choose the unknown while keeping the same physical quantities
Unknown	Rearrangement	Result
Stock aliquot volume	$V_1=\frac{c_2V_2}{c_1}$	$6.25\,\mathrm{mL}$
Final concentration	$c_2=\frac{c_1V_1}{V_2}$	$25.0\,\mathrm{mM}$
Final solution volume	$V_2=\frac{c_1V_1}{c_2}$	$1.00\times10^{2}\,\mathrm{mL}$
Stock concentration	$c_1=\frac{c_2V_2}{V_1}$	$0.400\,\mathrm{mol/L}$

The table reports each solved result to three significant figures, matching the least precise inputs for that rearrangement. The calculator retains guard digits and can display more. Its display setting does not establish measurement uncertainty or justify extra reported precision.

A positive target above the stock concentration cannot be reached merely by adding solvent. A zero stock cannot supply a positive amount of solute, and a zero final solution volume is not a valid molarity denominator. Check these limits before interpreting an algebraic answer as a feasible dilution.

Serial dilution multiplies stage factors

Start with the $25.0\,\mathrm{mM}$ solution from the preceding example. Model three intended tenfold stages, each using a $1.00\,\mathrm{mL}$ aliquot from the preceding homogeneous solution and a $10.00\,\mathrm{mL}$ final stage volume. Conservation applies within each aliquot dilution; only an aliquot, not all solute in a container, continues to the next stage.

Equal stage factors multiply

The vertical concentration scale is logarithmic: each equal downward step divides concentration by ten. Three stages take $25.0\,\mathrm{mM}$ to $0.0250\,\mathrm{mM}=25.0\,\mathrm{\mu M}$ . This does not show equal absolute decreases or physical container sizes.

1. Calculate the stage factor

D=\frac{V_{\mathrm{final}}}{V_{\mathrm{aliquot}}}=\frac{10.00}{1.00}=10

2. Apply the factor to every stage

c_k=\frac{c_0}{D^k}

c_3=\frac{25.0}{10^3}\,\mathrm{mM}=0.0250\,\mathrm{mM}

The final concentration is $25.0\,\mathrm{\mu M}$ . The total factor is $1000$ , not the sum of three factors. With zero stages, $D^0=1$ and the starting concentration is unchanged.

3. Track the stage concentrations and retained volume

Each nominal stage uses the same aliquot and final-volume ratio
Stage	Concentration	Cumulative factor	Retained after next transfer
$1$	$2.50\,\mathrm{mM}$	$10$	$9.00\,\mathrm{mL}$
$2$	$0.250\,\mathrm{mM}$	$100$	$9.00\,\mathrm{mL}$
$3$	$0.0250\,\mathrm{mM}$	$1000$	$10.00\,\mathrm{mL}$

Removing the next well-mixed aliquot leaves $9.00\,\mathrm{mL}$ at each of the first two stages without changing its concentration. The final stage has no outgoing transfer in this plan and retains $10.00\,\mathrm{mL}$ . The table does not propagate transfer, calibration or mixing uncertainties.

Inspect the serial plan and its direct-dilution comparison. The tool also shows a single-step comparison: $0.0100\,\mathrm{mL}=10.0\,\mathrm{\mu L}$ of the starting stock for the same $10.00\,\mathrm{mL}$ final target. That direct aliquot differs from the $1.00\,\mathrm{mL}$ per-stage aliquots. Read the labeled serial rows rather than confusing the two methods.

Percentages, ppm and molality need different denominators

Concentration labels describe a basis as well as a scale. OpenStax's concentration-unit chapter distinguishes the common percentage and parts-per-million conventions. A label that says only “percent” or “ppm” is incomplete when its basis is not otherwise specified.

Mass percent compares two masses

Consider a separate hypothetical solution containing $12.0\,\mathrm g$ of solute in $300.0\,\mathrm g$ of total solution. The denominator includes both solute and solvent:

w_{\mathrm{solute}}=\frac{12.0\,\mathrm g}{300.0\,\mathrm g}=0.0400

100w_{\mathrm{solute}}\,\%=4.00\,\%\,(\mathrm{m/m})

Mass/volume percent uses solution volume

If that separate solution has volume $250.0\,\mathrm{mL}$ , its mass/volume label follows a different ratio:

\frac{12.0\,\mathrm g}{250.0\,\mathrm{mL}}=\frac{4.80\,\mathrm g}{100\,\mathrm{mL}}

This is conventionally reported as $4.80\,\%\,(\mathrm{m/v})$ . It differs from the mass percent because the denominators differ. Labels also commonly use $\mathrm{w/w}$ and $\mathrm{w/v}$ for these two bases. One mass/volume percent means one gram per one hundred milliliters, so $1.00\,\%\,(\mathrm{m/v})=10.0\,\mathrm{g/L}$ . Check the stated mass/volume conversion.

Volume percent conventionally refers to component volume per final solution volume. An ideal volume-fraction dilution calculation needs compatible volume bases; it is not a mass percentage. Composition and density cannot be inferred from a bare percentage label.

Mass-based ppm is not universally milligrams per liter

A mass fraction of $10^{-6}$ corresponds to $1\,\mathrm{mg/kg}$ , conventionally one part per million by mass. To convert a mass-per-volume result, use the solution density. For a separate example with $c_{\mathrm{mass}}=2.00\,\mathrm{mg/L}$ and $\rho_{\mathrm{solution}}=1.20\,\mathrm{kg/L}$ :

\frac{c_{\mathrm{mass}}}{\rho_{\mathrm{solution}}}=\frac{2.00\,\mathrm{mg/L}}{1.20\,\mathrm{kg/L}}\approx1.67\,\mathrm{mg/kg}

The result is approximately $1.67$ ppm by mass. Its numerical value would match the milligrams-per-liter value only under a density of $1\,\mathrm{kg/L}$ . The familiar dilute-water approximation is not an exact general conversion. The Density Calculator helps check a separately known mass-volume relationship.

Molality uses solvent mass

b=\frac{n_{\mathrm{solute}}}{m_{\mathrm{solvent}}}

b=\frac{1.00\,\mathrm{mol}}{0.500\,\mathrm{kg}}=2.00\,\mathrm{mol/kg}

That solvent-mass example does not determine molarity without a solution volume or sufficient additional data. OpenStax's molality discussion explains why a volume-dependent concentration and a mass-based concentration respond differently to thermal expansion when composition remains fixed.

pH needs an activity or an explicit concentration model

Formal pH uses dimensionless hydrogen ion activity, $a_{\mathrm{H}^{+}}$ . A total analytical salt concentration does not supply that activity. In a dilute ideal aqueous approximation, a known equilibrium hydrogen ion concentration can be normalized by $c^\circ=1\,\mathrm{mol/L}$ :

\mathrm{pH}=-\log_{10}a_{\mathrm{H}^{+}}

\mathrm{pH}\approx-\log_{10}\!\left(\frac{[\mathrm{H}^{+}]}{c^\circ}\right)

The first is the activity definition in the IUPAC Gold Book; the second is the model used for the following calculation. Normalization makes the logarithm's argument dimensionless. Nonideal activity effects and actual pH measurement are additional questions.

Equal pH steps represent tenfold changes

The vertical ion-concentration scale is logarithmic. Going from $\mathrm{pH}=4$ to $\mathrm{pH}=7$ reduces the modeled hydrogen ion concentration by $10^3=1000$ . Neutrality at the last point applies to the adopted $\mathrm pK_{\mathrm w}=14.00$ model; the plot does not infer pH from total salt concentration.

Known hydrogen ion concentration

[\mathrm{H}^{+}]=10^{-4}\,\mathrm{mol/L}

\mathrm{pH}\approx-\log_{10}(10^{-4})=4.0000

With the adopted $\mathrm pK_{\mathrm w}=14.00$ water model, $\mathrm{pOH}=10.0000$ and $[\mathrm{OH}^{-}]=10^{-10}\,\mathrm{mol/L}$ . The decimal places here illustrate the calculator display; they do not establish a measurement uncertainty.

Check the known-ion conversion

Neutrality follows the water model

\mathrm{pH}+\mathrm{pOH}=\mathrm pK_{\mathrm w}

\mathrm{pH}_{\mathrm{neutral}}=\frac{\mathrm pK_{\mathrm w}}{2}

At approximately $25\,{}^\circ\mathrm C$ , the familiar model adopts $\mathrm pK_{\mathrm w}=14.00$ , giving neutral $\mathrm{pH}=7.00$ . Check the standard neutral reference. If a separate illustrative model adopts $\mathrm pK_{\mathrm w}=13.60$ , neutral pH is $6.80$ instead. That custom constant does not imply an automatic temperature lookup. Inspect the custom-water example. OpenStax explains the temperature dependence.

Analytical acid concentration is a different input

For an ideal fully dissociated monoprotic acid with analytical concentration $10^{-8}\,\mathrm{mol/L}$ , ignoring water would incorrectly suggest $\mathrm{pH}=8$ . Include water ionization and let $h=[\mathrm{H}^{+}]/c^\circ$ , $a=C_{\mathrm{acid}}/c^\circ$ and $K_{\mathrm w}=10^{-14}$ in the normalized model:

h=\frac{a+\sqrt{a^2+4K_{\mathrm w}}}{2}

\mathrm{pH}\approx-\log_{10}h\approx6.9783

The positive root approaches the neutral reference from the acidic side as the acid concentration approaches zero. This example assumes complete dissociation and no other acid-base equilibria or activity corrections. Compare the water-aware analytical-acid calculation. Weak acids, buffers and other mixtures need their corresponding equilibrium models; total solute molarity alone is insufficient.

Audit the result before reporting it

Identify the species and basis. Distinguish analytical solute amount, equilibrium ion concentration and activity.
Match the material. Check chemical formula, hydrate form, adopted molar mass and any supplied purity information.
Normalize the units. Keep concentration prefixes and volume conversions visible.
Check conservation. Compare solute amount before and after each aliquot dilution.
Test the direction. Adding solvent with no added solute should not raise the concentration of that solute.
Keep guard digits. Round the final report under a stated convention; calculator display settings do not propagate physical uncertainty.

The Lab Report Calculations guide expands the reporting and uncertainty distinction. The Significant Figures Calculator checks a stated classroom rounding rule, and the Common Unit Conversions guide helps keep dimensional conversions separate from changes in composition.

Practice with the basis written explicitly

Try each question, then open its worked answer. All values are hypothetical and use the stated nominal model.

Practice: amount and final solution volume

A final solution volume of $300.0\,\mathrm{mL}$ contains $0.0150\,\mathrm{mol}$ of a specified solute. Find its molarity and millimolar concentration.

c=\frac{0.0150}{0.3000}\,\mathrm{mol/L}=0.0500\,\mathrm{mol/L}

c=50.0\,\mathrm{mM}

The volume is converted to liters first. Three significant figures are retained under the classroom division rule.

Practice: choose the stock aliquot

A stock is $100.0\,\mathrm{mM}$ . Find the aliquot represented in $200.0\,\mathrm{mL}$ of a $5.00\,\mathrm{mM}$ target, assuming solute conservation.

V_1=\frac{(5.00)(200.0)}{100.0}\,\mathrm{mL}=10.0\,\mathrm{mL}

Matching concentration units cancel. The target is lower than the stock, so the final volume must exceed the aliquot volume.

Practice: multiply serial stage factors

Start with $80.0\,\mathrm{\mu M}$ and apply three intended fourfold dilutions. What is the final nominal concentration?

c_3=\frac{80.0}{4^3}\,\mathrm{\mu M}=1.25\,\mathrm{\mu M}

The overall factor is the product of the three stage factors. It does not describe a subtraction repeated three times.

FAQ

Frequently Asked Questions

What is molarity, and how is it different from molar mass?

Molarity is amount of a specified solute per final solution volume, commonly expressed in

\mathrm{mol/L}

. Molar mass is mass per amount, commonly expressed in

\mathrm{g/mol}

. They answer different questions: use molar mass to convert a weighed sample to moles, then divide those moles by solution volume. In this guide

c

denotes molarity and

M_{\mathrm m}

denotes molar mass, avoiding an ambiguous reuse of the same symbol.

Can I use the same molar mass for an anhydrous compound and its hydrate?

Use the formula and composition of the material actually specified. A hydrate includes water of crystallization, so its molar mass differs from the anhydrous form. An assay or purity correction may also change how much of the weighed sample represents the intended solute. The sodium chloride example assumes pure anhydrous material and adopted nominal atomic weights; it does not silently correct for a hydrate, moisture or impurities.

Why does the final solution volume differ from the solvent volume added?

Concentration uses the volume of the complete solution. Dissolving or mixing can change volume, so two measured liquid volumes need not add exactly. In our dilution,

100.0\,\mathrm{mL}

names the intended final solution volume; the arithmetic difference of

93.75\,\mathrm{mL}

is a nominal volume difference, not a guaranteed separate water addition. The calculation identifies quantities, while an appropriate measurement procedure establishes the final volume.

When can I use the dilution equation?

Use

c_1V_1=c_2V_2

when the same amount of the specified solute is conserved and the two concentrations are expressed on compatible volume-based bases. It does not account for reaction, precipitation, loss of solute or an added amount of solute. Mass percent and molality have different denominators, so their numerical values cannot simply replace molarity in that equation. A target above the stock requires concentration rather than adding solvent.

How do serial dilution factors combine?

Multiply the stage factors. Three intended tenfold stages give an overall factor of

10^3=1000

, so a starting concentration of

25.0\,\mathrm{mM}

becomes

0.0250\,\mathrm{mM}

in the ideal calculation. Each stage uses an aliquot from the preceding well-mixed stage. Zero stages leaves the starting concentration unchanged. Actual transfer and volume uncertainties still matter; exact arithmetic on the planned ratio does not make a physical preparation exact.

Are mass percent, mass/volume percent and ppm interchangeable?

Each label needs its denominator. Mass percent compares solute mass with total solution mass; mass/volume percent conventionally gives grams per one hundred milliliters of solution. Parts per million also needs a stated basis. Mass-based ppm is numerically milligrams per kilogram, whereas milligrams per liter requires solution density for conversion. Approximating their numerical equality for dilute water is an assumption, not a definition or a universal conversion.

Does knowing a solution's molarity tell me its pH?

No. A total analytical solute concentration does not identify the equilibrium hydrogen ion activity. Dissociation, acid-base equilibria, other solutes, water ionization and nonideal behavior may matter. Formal pH is

-\log_{10}a_{\mathrm{H}^{+}}

; this guide's simple concentration examples assume an ideal dilute aqueous model. A known equilibrium hydrogen ion concentration is a different input from an analytical acid concentration, especially near neutral conditions.

Is neutral pH always seven, and do displayed digits show uncertainty?

In the ideal water model, neutral pH is

\mathrm pK_{\mathrm w}/2

. It is

7.00

when the adopted water constant has

\mathrm pK_{\mathrm w}=14.00

, approximately the common aqueous reference at

25\,{}^\circ\mathrm C

. Different conditions can change that constant. The calculator's decimal places control display, not measurement uncertainty. Keep the temperature, model, original readings and reporting convention visible when interpreting an answer.

Sources and verification method

The primary references below support definitions and chemistry assumptions. Worked inputs are original hypothetical examples. Exact rational arithmetic independently checks molar mass, amount, unit conversions and dilution conservation; high-precision logarithms and the positive water-equilibrium root check the pH examples. Figure coordinates are derived from those models and checked against the rendered SVG artists. Exact nominal arithmetic is distinct from physical uncertainty.

OpenStax Chemistry 2e: molarity and dilution
Amount per final solution volume and conservation of a solute during dilution.
CIAAW: Abridged Standard Atomic Weights 2024
The adopted nominal sodium and chlorine weights used in the molar-mass example; their physical uncertainty is not propagated here.
BIPM: SI base unit mole
Amount of substance, specified entities and the SI definition of the mole.
OpenStax Chemistry 2e: other concentration units
Mass, volume and mass/volume percentages; the basis behind parts-per-million notation.
OpenStax Chemistry 2e: molality and mole fraction
Molality uses solvent mass; molarity uses solution volume and therefore depends on volume changes with temperature.
IUPAC Gold Book: pH
The activity-based definition, distinct from a dilute-solution concentration approximation.
OpenStax Chemistry 2e: pH and pOH
Logarithmic ion concentrations, the water constant and temperature-dependent neutrality.

Editorial Transparency

Who maintains this page

Ownership and review

Written and maintained by the Toolarithm editorial team. The method was last reviewed on October 30, 2026. No independent professional review is claimed.

Methodology

The editorial team checks definitions, examples, linked tools, and cited references. Dates shown here come from the page's maintained publication record.

Read the editorial methodology

Dates and sources

Published June 10, 2026. Updated October 30, 2026.