Ionic EquilibriumHard

Question

The range of most suitable indicator which should be used for titration of NaX (0.1 M, 10 ml) with 0.1 M HCl should be (Kb of X− = 10−6)

Options

A.2 – 3
B.3 – 5
C.6 – 8
D.8 –10

Solution

$X^{-} + H^{+} \rightleftharpoons HX$

$\frac{10 \times 0.1}{20}M\frac{10 \times 0.1}{20}M0$

Final 0 0 0.05 M

(10 ml HCl solution is needed for equivalent point)

At equivalent point [H+] = $\sqrt{K_{a} \times C}$

$= \sqrt{10^{- 6} \times 0.05}$

$\therefore P^{H} = - {\log\left( 5 \times 10^{- 8} \right)}^{1/2} = 3.65$

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