# Finding the pH of a mixture of weak acid and strong base

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?

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

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:

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

or

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

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.