The value of K_{sp} for B_2CO_3 is 3.2 times | vedclass

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

Question

(D) The dissociation of $B_2CO_3$ is given by: $B_2CO_3(s) \rightleftharpoons 2B^+(aq) + CO_3^{2-}(aq)$.Let the solubility of $B_2CO_3$ be $S \ M$.The

Vedclass · Accepted Answer

$4 	imes 10^{-2} \ M$

Vedclass · Answer

(D) The dissociation of $B_2CO_3$ is given by: $B_2CO_3(s) ightleftharpoons 2B^+(aq) + CO_3^{2-}(aq)$.Let the solubility of $B_2CO_3$ be $S \ M$.Then,$[B^+] = 2S$ and $[CO_3^{2-}] = S$.The solubility product expression is: $K_{sp} = [B^+]^2 [CO_3^{2-}] = (2S)^2(S) = 4S^3$.Given $K_{sp} = 3.2 	imes 10^{-5}$,we have $4S^3 = 3.2 	imes 10^{-5}$.$S^3 = 0.8 	imes 10^{-5} = 8 	imes 10^{-6}$.Taking the cube root,$S = 2 	imes 10^{-2} \ M$.The concentration of $B^+$ is $[B^+] = 2S = 2(2 	imes 10^{-2} \ M) = 4 	imes 10^{-2} \ M$.

The value of $K_{sp}$ for $B_2CO_3$ is $3.2 \times 10^{-5}$. The concentration of $B^{+}$ in its saturated solution is :-

Similar Questions

If the solubility product $(K_{sp})$ of a sparingly soluble salt $MX_2$ at $25 \, ^\circ C$ is $1.0 \times 10^{-11}$,what is the solubility of the salt in $mol \, L^{-1}$ at this temperature?

Zinc carbonate is precipitated from zinc sulphate solution by the addition of

The precipitate of $CaF_2$ $(K_{sp} = 1.7 \times 10^{-10})$ is obtained when equal volumes of the following are mixed:

Which of the following equations represents the relation between solubility and solubility product for salt $BA_3$?

Assertion : $BaCO_3$ is more soluble in $HNO_3$ than in plain water.
Reason : Carbonate is a weak base and reacts with the $H^+$ from the strong acid,causing the barium salt to dissociate.

Vedclass Test Series

Exam Paper Generator

Online Exam Module