Chemistry - Ionic Equilibrium Question with Solution | TestHub

pH of buffer is given by:

$pH=-\log K_a+\log \frac{[\text{ Conjugate base }]}{[\text{ Acid }]}$

Case-I: $4=-\log 1.0\times 10^{-5}+\log \frac{[\text{ Conjugate base }]}{(0.5)}$

$\therefore \log \frac{[\text{ Conjugate base }]}{0.5}=-1$

[Conjugate base] $=0.1\times 0.5=0.05M$

Case-II: $6=-\log 1.0\times 10^{-5}+\log \frac{\text{ [Conjugate base] }}{0.5}$

$\therefore \log \frac{[\text{ Conjugate base }]}{0.5}=1$

∴ [Conjugate base $]=[$ Salt $]=10\times 0.5=5\mathrm{M}$

Now the two buffer $[1.\mathrm{NaA}=0.05\mathrm{M}$ and $HA=0.5M$ and $(II.\mathrm{NaA}=5M$ and $HA=0.5M)]$ are mixed in equal proportion.

Thus, new conc. of $NaA$ is mixed buffer $=\frac{0.05\times V+5\times V}{2V}=\frac{5.05}{2}$

New conc. of HA in mixed buffer $=\frac{0.5\times V+0.5\times V}{2V}=0.5\mathrm{M}$

Thus, $pH=-\log 1.0\times 10^{-5}+\log \frac{[5.05/2]}{[0.5]}$

$$\begin{array}{rlr}pH=5+0.7033=5.7033& & \end{array}$$

$$2{\mathrm{N}}_2{\mathrm{O}}_5(g)\to 4{\mathrm{NO}}_2(g)+{\mathrm{O}}_2(g)$$