5.1.3(d)-(f) - pH, Kw and strong acid-base calculations

pH as shorthand for hydrogen ion concentration

In aqueous acid-base chemistry, $[\ce{H+}]$ means the concentration of hydrogen ions in $\,\mathrm{mol} \,\mathrm{dm}^{-3}$. The concentration can be tiny: neutral water at 298 K has $[\ce{H+}] = 1.00 \times 10^{-7} \,\mathrm{mol} \,\mathrm{dm}^{-3}$. pH turns that awkward number into a convenient value.

The log is base 10. A change of 1 pH unit means a factor of 10 change in $[\ce{H+}]$. A solution with pH 2 has ten times the $[\ce{H+}]$ of a solution with pH 3, and one hundred times the $[\ce{H+}]$ of a solution with pH 4.

The minus sign is there because $[\ce{H+}]$ is usually less than 1. For example:

• if $[\ce{H+}] = 1.00 \times 10^{-2} \,\mathrm{mol} \,\mathrm{dm}^{-3}$, then $\mathrm{pH} = 2.00$
• if $[\ce{H+}] = 1.00 \times 10^{-5} \,\mathrm{mol} \,\mathrm{dm}^{-3}$, then $\mathrm{pH} = 5.00$

So lower pH means higher $[\ce{H+}]$. This is the HSW8 point in the OCR specification: pH is useful terminology because it communicates hydrogen ion concentration on a manageable scale.

Finding hydrogen ion concentration from pH

The pH equation can be reversed. If you know pH, use:

This is where calculator handling matters. Use the $10^{x}$ or inverse-log function, not multiplication by 10.

Worked example: calculate $[\ce{H+}]$ for a solution with pH 2.37.

1. Write the expression:
   $[\ce{H+}] = 10^{-\mathrm{pH}}$
2. Substitute:
   $[\ce{H+}] = 10^{-2.37}$
3. Calculate: $$[\ce{H+}] = 4.27 \times 10^{-3} \,\mathrm{mol} \,\mathrm{dm}^{-3}$$

The answer is sensible: pH 2.37 is between pH 2 and pH 3, so $[\ce{H+}]$ should be between $1 \times 10^{-2}$ and $1 \times 10^{-3} \,\mathrm{mol} \,\mathrm{dm}^{-3}$.

Do not round too early in a longer calculation. Keep the full calculator value until the final line, then round to the precision requested by the question.

$K_w$ and neutral water

Water contains a very small concentration of hydrogen ions and hydroxide ions. The ionic product of water links the two concentrations:

On the OCR data sheet at 298 K:

The units come from multiplying two concentrations, each in $\,\mathrm{mol} \,\mathrm{dm}^{-3}$.

In neutral water, the concentrations are equal:

$[\ce{H+}] = [\ce{OH-}]$

At 298 K:

So:

and:

The important idea is not "neutral always means pH 7". Neutral means $[\ce{H+}] = [\ce{OH-}]$. If a question gives a different value of $K_w$ at a different temperature, use the value given in the question.

Worked example

The value of $K_w$ for water at 40 °C is $2.92 \times 10^{-14} \,\mathrm{mol}^{2} \,\mathrm{dm}^{-6}$. Calculate the pH of neutral water at this temperature.

1. For neutral water, $[\ce{H+}] = [\ce{OH-}]$.
2. So $K_w = [\ce{H+}]^{2}$.
3. $[\ce{H+}] = \sqrt{2.92 \times 10^{-14}} = 1.71 \times 10^{-7} \,\mathrm{mol} \,\mathrm{dm}^{-3}$.
4. $\mathrm{pH} = -\log(1.71 \times 10^{-7}) = 6.77$.

This water is neutral because $[\ce{H+}] = [\ce{OH-}]$, even though its pH is below 7.

Strong monobasic acid calculations

A strong acid is fully dissociated in aqueous solution. A monobasic acid donates one proton per acid molecule. For a strong monobasic acid such as hydrochloric acid, $\ce{HCl}$, or nitric acid, $\ce{HNO3}$:

$\text{acid concentration} = [\ce{H+}]$

This shortcut only works because both words matter:

• strong: complete dissociation
• monobasic: one $\ce{H+}$ per acid molecule

Strength is not the same as concentration. A strong acid is one that dissociates completely; a concentrated acid is one with many moles of acid per $\,\mathrm{dm}^{3}$. OCR calculations can use dilute strong acids, so always use the concentration given in the question.

Worked example: calculate the pH of $0.0350 \,\mathrm{mol} \,\mathrm{dm}^{-3} \ce{HCl(aq)}$.

Because $\ce{HCl}$ is a strong monobasic acid: