# Determination of pKb of a mono acidic base

$$20$$ 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,

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

Where,
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

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

• 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 – MaxW Mar 9 at 17:48
• @MaxW ok sir , thanks for upvoting , but will it not form a buffer ?? – Advil Sell Mar 9 at 17:54