Concentration of H_2SO_4 and Na_2SO_4 in a so | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) $H_2SO_4 \rightarrow H^{+} + HSO_4^{-}$$Na_2SO_4 \rightarrow 2Na^{+} + SO_4^{2-}$$HSO_4^{-} \rightleftharpoons H^{+} + SO_4^{2-}; \ K_{a2} = 1.2 \

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(D) $H_2SO_4 ightarrow H^{+} + HSO_4^{-}$$Na_2SO_4 ightarrow 2Na^{+} + SO_4^{2-}$$HSO_4^{-} ightleftharpoons H^{+} + SO_4^{2-}; \ K_{a2} = 1.2 	imes 10^{-2} \ M$Let the concentration of $SO_4^{2-}$ be $[SO_4^{2-}] = 1.8 	imes 10^{-2} - x$,where $x$ is the amount of $SO_4^{2-}$ consumed by $H^{+}$ to form $HSO_4^{-}$.$K_{a2} = \frac{[H^{+}][SO_4^{2-}]}{[HSO_4^{-}]} = \frac{(1+x)(1.8 	imes 10^{-2} - x)}{(1-x)} = 1.2 	imes 10^{-2}$Assuming $x$ is small,$1+x \approx 1$ and $1-x \approx 1$,so $1.8 	imes 10^{-2} - x = 1.2 	imes 10^{-2} \Rightarrow x = 0.6 	imes 10^{-2} \ M$.Thus,$[SO_4^{2-}] = 1.8 	imes 10^{-2} - 0.6 	imes 10^{-2} = 1.2 	imes 10^{-2} \ M$.For $PbSO_4 ightleftharpoons Pb^{2+} + SO_4^{2-}$,$K_{sp} = [Pb^{2+}][SO_4^{2-}] = s(s + 1.2 	imes 10^{-2}) = 1.6 	imes 10^{-8}$.Since $s$ is very small,$s + 1.2 	imes 10^{-2} \approx 1.2 	imes 10^{-2}$.$s(1.2 	imes 10^{-2}) = 1.6 	imes 10^{-8} \Rightarrow s = \frac{1.6}{1.2} 	imes 10^{-6} = 1.33 	imes 10^{-6} \ M$.Comparing with $X 	imes 10^{-Y} \ M$,we get $Y = 6$.

Similar Questions

The solubility product of a sparingly soluble salt $AX_2$ is $3.2 \times 10^{-11}$. Its solubility in $mol/L$ is:

Calculate the ionic concentration of a sparingly soluble salt $BA$ in $mol \ dm^{-3}$ at $300 \ K$ when equilibrium is attained,if the solubility product of the salt is $2.7 \times 10^{-10}$ at the same temperature.

The solubility of an $MX_2$ type electrolyte is $0.5 \times 10^{-4} \ mol/L$. The value of $K_{sp}$ of the electrolyte is:

If the solubility product $(K_{sp})$ of $Ni(OH)_2$ is $1.9 \times 10^{-15}$,the molar solubility of $Ni(OH)_2$ in $1.0 \ M \ NaOH$ is:

Explain the difference between ionic product and solubility product. Explain the condition for the precipitation of a sparingly soluble salt.

Vedclass Test Series

Exam Paper Generator

Online Exam Module