Chemistry - Ionic Equilibrium Question with Solution | TestHub

$\because \mathrm{pH}=4.7$

∴ resulting solution is buffer.

$$\begin{array}{rlrlrl}& \therefore \mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}+\log \left(\frac{{\mathrm{n}}_{{\mathrm{A}}^{-}}}{{\mathrm{n}}_{\mathrm{HA}}}\right)& & & & \end{array}$$

$4.7=-\log \left(\frac{10^{-14}}{5\times 10^{-10}}\right)+\log \left(\frac{{\mathrm{n}}_{{\mathrm{A}}^{-}}}{{\mathrm{n}}_{\mathrm{HA}}}\right)$

$4.7=4.7+\log \left(\frac{{\mathrm{n}}_{{\mathrm{A}}^{-}}}{{\mathrm{n}}_{\mathrm{HA}}}\right)$

$\log \left(\frac{{\mathrm{n}}_{{\mathrm{A}}^{-}}}{{\mathrm{n}}_{\mathrm{HA}}}\right)=0$

$$\therefore {\mathrm{n}}_{{\mathrm{A}}^{-}}={\mathrm{n}}_{\mathrm{HA}}(finally)$$

Initially : m.m. of HA= $0.1\times 100=10\mathrm{m}.\mathrm{m}$.

Finally : m.m. of H.A. $=\mathrm{m}.\mathrm{m}$. of ${\mathrm{A}}^{-}=5\mathrm{m}.\mathrm{m}$.

∴ Total m.m. of NaOH used $=5\mathrm{m}.\mathrm{m}$.

$$\begin{array}{rlr}& (0.2\times 22.5)+(0.1\times V)=5& \end{array}$$

$$V=5\mathrm{ml}$$