12.11-12.13 - Kw, pKa And Strong-Base pH

Water Sets The pH Scale

A bottle labelled "pure water" is not a bottle with absolutely no ions. A tiny fraction of water molecules form hydrogen ions and hydroxide ions, and at a fixed temperature those ion concentrations are linked by an equilibrium expression.

The simplified reversible change is:

$\ce{H2O(l) <=> H+(aq) + OH-(aq)}$

In reality, $\ce{H+(aq)}$ is a shorthand for hydrated hydrogen ions in water. $\mathrm{pH}$ calculations use the shorthand $[\ce{H+}]$, so the important expression is:

[DIAGRAM: asset_name: Water Ion Product Particles; asset_slug: edexcel_a_level_chemistry_l055_water_ion_product_particles; recommended_method: deterministic_chart; description: Pale-background colour beaker and particle zoom showing many water molecules with rare positive and negative ions to support the ionic product of water model.]

At 25 ${}^{\circ}\mathrm C$ (298 K), the data value used unless another value is supplied is:

In pure water, the two ions form in equal amounts, so:

That gives:

This is why neutral pure water has $\mathrm{pH}$ 7 under these conditions. Neutral does not mean "no ions"; it means equal hydrogen ion and hydroxide ion concentrations. At another temperature the neutral $\mathrm{pH}$ is $\mathrm pK_{\mathrm w}$/2, so neutral water need not have $\mathrm{pH}$ 7.

From Strong Base Concentration To pH

A strong base is treated as fully dissociated in aqueous solution. For sodium hydroxide:

$\ce{NaOH(aq) -> Na+(aq) + OH-(aq)}$

So, for $0.0250 \mathrm{mol\,dm^{-3}} \ce{NaOH}$, the hydroxide ion concentration is also:

The $\mathrm{pH}$ is not found by doing $-\log_{10}[\ce{OH-}]$. That would give $\mathrm{pOH}$. To calculate $\mathrm{pH}$ using $K_{\mathrm w}$, first find $[\ce{H+}]$.

Use this route:

1. Find $[\ce{OH-}]$ from the strong base concentration and formula.
2. Use $K_{\mathrm w} = [\ce{H+}][\ce{OH-}]$.
3. Rearrange: $[\ce{H+}] = K_{\mathrm w} / [\ce{OH-}]$.
4. Calculate $\mathrm{pH} = -\log_{10}[\ce{H+}]$.

Worked example: $\mathrm{pH}$ of sodium hydroxide

Calculate the $\mathrm{pH}$ of $0.0250 \mathrm{mol\,dm^{-3}} \ce{NaOH}$.

For sodium hydroxide, one formula unit gives one hydroxide ion:

Use $K_{\mathrm w}$:

$[\ce{H+}] = K_{\mathrm w} / [\ce{OH-}]$

Now calculate $\mathrm{pH}$:

$\mathrm{pH} = 12.40$

Keep the unrounded calculator value until the final logarithm. These strong-base routes assume the hydroxide concentration is large enough for water’s own ionisation to be negligible.

The chemical sense-check is that this is a strongly alkaline solution, so the $\mathrm{pH}$ should be above 7. The $[\ce{H+}]$ is tiny, but it is not zero.

Alternative check using $\mathrm{pOH}$

You may also see the route:

$\mathrm{pOH} = -\log_{10}[\ce{OH-}]$

$\mathrm{pH} + \mathrm{pOH} = 14.00$

For this example:

$\mathrm{pH} = 14.00 - 1.60 = 12.40$

That shortcut works here because $\mathrm pK_{\mathrm w} = 14.00$ for $K_{\mathrm w} = 1.00\times 10^{-14}$. The $K_{\mathrm w}$ route is safer because it shows where the relationship comes from.

Get Hydroxide Concentration First

Strong-base questions often hide one decision before the logarithm: the concentration of the base is not always the same as the concentration of hydroxide ions.

For a dissolved Group 1 hydroxide:

$\ce{NaOH(aq) -> Na+(aq) + OH-(aq)}$

$1\,\mathrm{mol} \ce{NaOH}$ gives $1\,\mathrm{mol} \ce{OH-}$.

For a dissolved Group 2 hydroxide:

$\ce{Ba(OH)2(aq) -> Ba2+(aq) + 2OH-(aq)}$

$1\,\mathrm{mol} \ce{Ba(OH)2}$ gives $2\,\mathrm{mol} \ce{OH-}$.

If a question gives the concentration of a strong base solution, read the formula before using $K_{\mathrm w}$.

Worked example: $\mathrm{pH}$ of barium hydroxide

Calculate the $\mathrm{pH}$ of $0.0150 \mathrm{mol\,dm^{-3}} \ce{Ba(OH)2}$.

First convert base concentration into hydroxide ion concentration:

Now use $K_{\mathrm w}$:

Then:

The answer is higher than for a $0.0150 \mathrm{mol\,dm^{-3}}$ one-hydroxide base because the solution contains twice as many hydroxide ions per mole of dissolved base.

Common-error contrast

Weak route: $\mathrm{pH} = -\log_{10}(0.0150) = 1.82$

That answer is acidic, so it fails the chemical sense-check. It also used the base concentration as if it were $[\ce{H+}]$.

Stronger route: find $[\ce{OH-}]$ from the base formula, use $K_{\mathrm w}$ to find $[\ce{H+}]$, then calculate $\mathrm{pH}$.

Dilution changes hydroxide concentration before any logarithm is taken. For example, 20.0 cm³ of a dissolved hydroxide made up to 100.0 cm³ has one fifth of its initial concentration. First apply $c_1V_1=c_2V_2$, then multiply by the number of hydroxide ions per formula unit.