pH = 3.44As

we know,

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

\( [H^{+}]\) = \( 3.63\times 10^{-4}\)

Now, \( K_{h} = \dfrac{3.63\times 10^{-4}}{0.02}\)

(Given that concentration = 0.02M)

\( \Rightarrow K_{h} = 6.6\times 10^{-6}\)

As we know that,

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

\( = \dfrac{K_{w}}{K_{h}} \)

\( = \dfrac{10^{-14}}{6.6\times 10^{-6}}\)

= \( 1.51×10^{−9}\)

Answered by Abhisek | 2 years 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?

What is the minimum volume of water required to dissolve 1g of calcium sulphate at 298 K? (For calcium sulphate, K_{sp} is 9.1 × 10^{–6}).

What is the maximum concentration of equimolar solutions of ferrous sulphate and sodium sulphide so that when mixed in equal volumes, there is no precipitation of iron sulphide? (For iron sulphide, Ksp = 6.3 × 10^{–18}).

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