$20$ mL of a weak monoacidic base($\text{BOH}$) requires $12$ mL of $0.3$ M $\text{HCl}$ solution for the equivalence point. During titration, the pH of the base solution was $10$ upon the addition of $4$ mL of $0.3$ M $\text{HCl}$ solution. What is the $\text{pKb}$ of the base($\text{BOH}$)?

My attempt:

The concentration of the base is given by,
$c * 20 = 12 * 0.3$ -- (equating the moles of the acid and base)
$c = 0.18$ M

So, initial moles of the base in the container $= 20 * 0.18 = 3.6$ mmol
Moles of acid that is added $ = 4 * 0.3 = 1.2$ mmol

Since the base is monoacidic, they will react in a 1:1-mole ratio. The acid is the limiting reagent, so it will be fully consumed. Therefore the moles of base left is,

$3.6 - 1.2 = 2.4$ mmol

The concentration of the base is,
$\frac{2.4}{24} = 0.1$ M

Since the pH of the solution is $10$, therefore the concentration of $\text{OH}^- = 10^{-4}$

Applying the approximated formula for calculating the $\text{k}_b$ of a weak base,

$$\text{K}_b = \frac{x^2}{c}$$

x = concentration of $\text{OH}^-$ ions
c = concetration of the base

In our case,
x = $10^{-4}$
c = $0.1$

Plugging in the values I got $pK_b = 7$. But the answer is given $4.3$.

Any help would be appreciated.


I appreciate your effort you showed in your question, but you missed one important thing that when a weak base is reacted with a strong acid a buffer solution is formed and the $\mathrm{pH}$ is not only contributed by base but also by the acidic salt ($\ce{BCl}$ in this case).

So, simply applying the Henderson–Hasselbach equation (derivation can be found here)

$$\mathrm{pOH} = \mathrm{p}K_\mathrm{b} + \log{\frac{[\ce{B-}]}{\ce{[BOH]}}}$$

we get

$$4 = \mathrm{p}K_\mathrm{b} + \log{\frac{1.2}{2.4}},$$

which gives us

$$\mathrm{p}K_\mathrm{b} = 4.3010$$

  • $\begingroup$ I upvoted because you got the math right, but your assertion that this is a buffer solution is beside the point. The real point here is that you can't assume that $\ce{[OH^-] \approx [B^+]}$ at pH=10.00 $\endgroup$ – MaxW Mar 9 at 17:48
  • $\begingroup$ @MaxW ok sir , thanks for upvoting , but will it not form a buffer ?? $\endgroup$ – Advil Sell Mar 9 at 17:54

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