1. Give the relationship between K_a, c and (' ') where 'K_a' is … - Vidyadip

Question

Give the relationship between $\text{K}_\text{a}$, c and $('\alpha')$ where '$\text{K}_\text{a}$' is acid dissociation constant, 'c’ is molar concentration, $'\alpha'$ is degree of dissociation.
If the solubility of $\mathrm{Ca}\left(\mathrm{lO}_3\right)_2$ in water at $18^{\circ} \mathrm{C}$ is $2.1 \mathrm{~g} /$ litre. Calculate the value of solubility product.

[Molecular mass of $\left.\mathrm{Ca}\left(\mathrm{lO}_3\right)_2=390\right]$

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Answer

$\text{K}_{\text{a}}=\frac{\text{c}\alpha^2}{1-\alpha}\text{ if }\alpha<<<<1\text{ then }\text{K}_{\text{a}}=\text{c}\alpha^2$

$\because1-\alpha=1$

Solubility in $\text{mol L}^{-1}$$=\frac{2.1\text{g L}^{-1}}{390\text{g mol}^{-1}}$

$=5.4\times10^{-3}\text{mol L}^{-1}$

$\text{Ca}(\text{IO}_3)_2\rightleftharpoons\text{Ca}^{2+}+2\text{IO}^-_3$

$\text{K}_{\text{sp}}=[\text{Ca}^{2+}][\text{IO}^-_3]^2$

$\text{K}_{\text{sp}}=(5.4\times10^{-3})(2\times5.4\times10^{-3})^2$

$\text{K}_{\text{sp}}=-629.856\times10^{-9}\text{mol}^3\text{L}^{-3}$

$=6.6\times10^{-7}\text{mol}^3\text{L}^{-3}$

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