Name: pH Scale
Uploaded: 2019-07-19
Duration: 9 min 45 s
Description: This video discusses the pH of acids, bases, and salts and also the limitations of using a pH scale for identification. The natural indicators do the verific...

Question

Derive expression for the calculation of

pH

of a solution of a salt of weak base with a strong acid.

Accepted Answer

For simplicity, let the salt be $BA$
$BA + H_{2} O ⇌ \underset{Weak Base}{BOH} + \underset{Strong Acid}{HA}$

Or, $B^{+} A^{-} + H_{2} O ⇌ BOH + H^{+} A^{-}$
Or, $B^{+} + H_{2} O ⇌ BOH + H^{+}$

The hydrolysis constant, $K_{h} = \frac{[BOH] [H^{+}]}{[B^{+}]} . . . . . . . . . . . . . . . . . . . . . . . (1)$

The dissociation constant $K_{b}$ for the weak base $BOH$ $= \frac{[B^{+}] [{OH}^{-}]}{[BOH]} . . . . . . . . . . . . . . (2)$

And, the ionic product of water, $K_{w} = [H^{+}] [{OH}^{-}] . . . . . . . . . . . . . . (3)$

Multiplying $(1), (2)$ and dividing by $(3)$ we get,
$\frac{K_{h} K_{b}}{K_{w}} = 1$ or $K_{h} = \frac{K_{w}}{K_{b}} . . . . . . . . . . . . . . (4)$

$B^{+} + H_{2} O ⇌ BOH + H^{+} Initial conc . C Conc . at Eq . C - Ch Ch Ch$

Here, $C$ is the initial concentration and $h$ is the degree of hydrolysis

The hydrolysis constant, $K_{h} = \frac{(Ch) (Ch)}{C (1 - h)}$
For a weak electrolyte, $h ≪ 1$

$∴ 1 - h \approx 1$

Hence, $K_{h} = {Ch}^{2}$
$\Rightarrow h = \sqrt{\frac{K_{h}}{C}}$

From $(4)$ and $(5)$

$h = \sqrt{\frac{K_{w}}{K_{b} C}} . . . . . . . . . . . . . . . . . . . . . . . . . . . (5)$

For $pH$ :

Now, $[H^{+}] = Ch$

$∴ [H^{+}] = C \sqrt{\frac{K_{w}}{K_{b} C}} = \sqrt{\frac{K_{w} C}{K_{b}}}$ or ${(\frac{K_{w} C}{K_{b}})}^{\frac{1}{2}}$

$∴ pH = - \log [H^{+}] = - \log {(\frac{K_{w} C}{K_{b}})}^{\frac{1}{2}} = - \frac{1}{2} \log (\frac{K_{w} C}{K_{b}}) = - \frac{1}{2} [\log K_{w} - \log K_{b} + \log C]$

Derive expression for the calculation of $pH$ of a solution of a salt of weak base with a strong acid.

Important Questions on Ionic Equilibrium in Solution

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