The K_{a} for chloroacetic acid (ClCH_{2}COOH) is \( 1.35\times 10^{-3}\).

\( \Rightarrow K_{a} = c\alpha ^{2} \)

\( \alpha = \sqrt{\dfrac{K_{a}}{c}} \)

\( = \sqrt{\dfrac{1.35\times 10^{-3}}{0.1}}\); (given concentration = 0.1M)

\( \alpha = \sqrt{1.35\times 10^{-2}} \)

= 0.116

\( [H^{+}] = c\alpha = 0.1×0.116 = 0.0116 \)

\( pH = -\log [H^{+}] = 1.94 \)

ClCH_{2}COONa is a salt of strong base i.e. NaOH, and weak acid i.e. ClCH_{2}COOH

\( ClCH_{2}COO^{-} + H_{2}O \)

\( \leftrightarrow ClCH_{2}COOH + OH^{-}\)

\( K_{h} = \dfrac{[ClCH_{2}COO][OH^{-}]}{[ClCH_{2}COO^{-}]} \)

Now, \( K_{h} = \dfrac{K_{w}}{K_{a}}\)

\( K_{h} = \dfrac{10^{-14}}{1.35\times 10^{-3}} \)

\( = 0.740\times 10^{-11}\)

Also, \( K_{h} = \dfrac{y^2}{0.1}\Rightarrow \; 0.740\times 10^{-11} \)

\( = \dfrac{y^2}{0.1}\Rightarrow \; 0.0740\times 10^{-11} = y^2\)

y =\( 0.86\times 10^{-6}\)

\( [OH^{-}] = 0.86\times 10^{-6}\)

\( [H^{+}] =\dfrac{K_{w}}{0.86\times 10^{-6}} \)

\( Kw =\dfrac{10^{-14}}{0.86\times 10^{-6}}\)

\( [H^{+}] = 1.162\times 10^{-3}\)

\( pH = - \log [H^{+}]= 7.94 \)

Answered by Abhisek | 1 year agoThe concentration of sulphide ion in 0.1M HCl solution saturated with hydrogen sulphide is 1.0 × 10^{–19} M. If 10 mL of this is added to 5 mL of 0.04M solution of the following: FeSO_{4}, MnCl_{2}, ZnCl_{2} and CdCl_{2} . in which of these solutions precipitation will take place?

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