If the solubility product of a base M(OH)_3 i | vedclass

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

Question

(B) The dissociation of the base is given by: $M(OH)_3 \rightleftharpoons M^{3+} + 3OH^-$Let the solubility be $s$. Then $[M^{3+}] = s$ and $[OH^-] =

Vedclass · Accepted Answer

$3 	imes 10^{-3}$

Vedclass · Answer

(B) The dissociation of the base is given by: $M(OH)_3 ightleftharpoons M^{3+} + 3OH^-$Let the solubility be $s$. Then $[M^{3+}] = s$ and $[OH^-] = 3s$.The solubility product expression is: $K_{sp} = [M^{3+}][OH^-]^3$Substituting the values: $2.7 	imes 10^{-11} = (s)(3s)^3$$2.7 	imes 10^{-11} = 27s^4$$s^4 = \frac{2.7 	imes 10^{-11}}{27} = 0.1 	imes 10^{-11} = 10^{-12}$$s = (10^{-12})^{1/4} = 10^{-3} \ M$The concentration of $OH^-$ is $[OH^-] = 3s = 3 	imes 10^{-3} \ M$.

If the solubility product of a base $M(OH)_3$ is $2.7 \times 10^{-11}$,find the concentration of $OH^-$.

Similar Questions

The solubility of $AgCl$ at $20\,^oC$ is $1.435 \times 10^{-3} \, g/L$. The solubility product of $AgCl$ is .......

$A$ solution is saturated with $SrCO_3$ and $SrF_2$. The $[CO_3^{2-}]$ is found to be $1.2 \times 10^{-3} \ M$. The concentration of $F^{-}$ in the solution would be:
Given: $K_{sp}(SrCO_3) = 7.0 \times 10^{-10}$, $K_{sp}(SrF_2) = 7.9 \times 10^{-10}$

If $K_{sp}$ is the solubility product of $Al(OH)_3$,its solubility is expressed by the formula:

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,$K_{sp} = 6.3 \times 10^{-18}$).

The solubility product of a salt having the general formula $MX_2$ in water is $4 \times 10^{-12}$. What is the concentration of $M^{2+}$ ions in the saturated aqueous solution of the salt?

Vedclass Test Series

Exam Paper Generator

Online Exam Module