What are the steps we go about in finding the pH of a solution of strong base and weak acid?

Here is the question that I've been given

What is the pH of a solution made by mixing $\pu{50 ml}$ of $\pu{0.2 M}$ $\ce{NH4Cl}$ and $\pu{75 ml}$ of $\pu{0.1 M}$ $\ce{NaOH}$, when $\mathrm pK_\mathrm b(\ce{NH3}) = 4.74$?

A. 7.02
B. 13.0
C. 9.73
D. 6.31

What I did to solve it was to use the Henderson equation for buffers, $$\begin{align} \mathrm{pOH} &= \mathrm pK_\mathrm b + \log{\frac{[\text{salt}]}{[\text{base}]}} \\ &= 4.74 + \log{\left(\frac{0.2 \cdot 50}{0.1 \cdot 75}\right)} \\ &= 4.74 + \log{\frac{4}{3}} \\ &= 4.86 \end{align}$$

and thus $\mathrm{pH} = 14 - \mathrm{pOH} = 9.14.$

The correct answer is C. But on putting values I'm getting the wrong answer. Can you help me figure out why? Also, shouldn't the buffer solution have a common ion?

  • 3
    $\begingroup$ What is the reaction between $\ce{NH4+}$ a weak acid, and $\ce{NaOH}$ a strong base? $\endgroup$
    – MaxW
    Mar 31, 2020 at 17:57

2 Answers 2


I admit Karsten Theis has given en excellent answer for OP's question. However, I'd like to point out that this could be solve without getting confused by $\mathrm{p}K_\mathrm{b}$, which is common with novices when using the Henderson–Hasselbalch equation for buffers. The equation is derived by dissociation of weak acid ($\ce{HA}$): $$\ce{HA + H2O <=> H3O+ + A-}$$ Hence, we can derived Henderson–Hasselbalch equation getting log value of $\mathrm{p}K_\mathrm{a} = \frac{[\ce{H3O+}][\ce{A-}]}{[\ce{AH}]}$ in both side and simplifying it as: $$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log{\frac{[\ce{A-}]}{[\ce{AH}]}}$$

This works fine with any buffer solution made with a weak acid and its conjugate base. However, most novices get confused by when the buffer made with a weak base and its conjugate acid. The confusion directed mainly by two fact:

  1. The weak base is usually provided by its $\mathrm{p}K_\mathrm{b}$ value (For example, $\mathrm{p}K_\mathrm{b}$ of ammonia is $\approx 4.3$ while $\mathrm{p}K_\mathrm{a}$ of ammonia is $\gt 34$, not $14 - \mathrm{p}K_\mathrm{b}$).
  2. The value of ($14 - \mathrm{p}K_\mathrm{b}$) is really belongs to conjugate base of ammonia, $\ce{NH4+}$ (made by the reaction with strong acid). It is a good practice that to use this value as $\mathrm{p}K_\mathrm{a}\mathrm{H}$ (See this article).

Accordingly, it is a good rule of thumb that we can use a equation for the dissociation of conjugate acid ($\ce{BH+}$) of a weak base: $$\ce{BH+ + H2O <=> H3O+ + B}$$

Thus, we can derived Henderson–Hasselbalch equation getting log value of $\mathrm{p}K_\mathrm{a}\mathrm{H} = \frac{[\ce{H3O+}][\ce{B}]}{[\ce{BH+}]}$ in both side and simplifying it as: $$\mathrm{pH} = \mathrm{p}K_\mathrm{a}\mathrm{H} + \log{\frac{[\ce{B}]}{[\ce{BH+}]}}$$

You reacted $\pu{0.010 mol}$ of ammonium salt (conjugate acid of a weak base) with $\pu{0.0075 mol}$ of $\ce{NaOH}$, a strong acid. It resulted $\pu{0.0075 mol}$ of ammonia (a weak base) and $\pu{0.0025 mmol}$ of unreacted ammonium salt in the solution, which is a buffer. Since both species in same volume, the ratio of weak base to weak acid, $\frac{[\ce{B}]}{[\ce{BH+}]}$ is $\frac{0.0075}{0.0025} =3$. Now, since $\mathrm{p}K_\mathrm{b}$ of ammonia is given as $4.74$, $\mathrm{p}K_\mathrm{a}\mathrm{H} = 14 - 4.74 = 9.26$.

If you substitute these values into above equation, you get the answer:

$$\mathrm{pH} = \mathrm{p}K_\mathrm{a}\mathrm{H} + \log{\frac{[\ce{B}]}{[\ce{BH+}]}} = 9.26 + \log 3 = 9.74$$

Note: Your error in calculation is miscalculation on $\frac{[\text{Base}]}{[\text{acid}]}$ ratio.


Buffer equation

The Henderson equation for buffers is:

$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log{\frac{[\ce{A-}]}{[\ce{AH}]}}$$

$\mathrm{p}K_\mathrm{a}$ and $\mathrm{p}K_\mathrm{b}$ add up to 14, as do $\mathrm{pH}$ and $\mathrm{pOH}$. So the expression for $\mathrm{pOH}$ is:

$$\mathrm{14 - pOH = 14} - \mathrm{p}K_\mathrm{b} + \log{\frac{[\ce{A-}]}{[\ce{AH}]}}$$


$$\mathrm{pOH} = \mathrm{p}K_\mathrm{b} - \log{\frac{[\ce{A-}]}{[\ce{AH}]}}$$

Amounts of weak acid and weak base

Also, shouldn't the buffer solution have a common ion?

You start with $\pu{10 mmol}$ of ammonium salt (weak acid), to which you add $\pu{7.5 mmol}$ of $\ce{NaOH}$. The result is $\pu{7.5 mmol}$ of ammonia (weak base) with $\pu{2.5 mmol}$ ammonium salt remaining. So the ratio of weak base to weak acid is 1:3.

If you plug this into one or the other buffer equation, you get the answer.


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