If the solubility product (K_{sp}) of Ni(OH)_ | vedclass

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

Question

(C) The dissociation of $NaOH$ is: $NaOH \rightarrow Na^+ + OH^-$. Since $NaOH$ is a strong electrolyte,$[OH^-] = 1.0 \ M$. Let the solubility of $Ni(

Vedclass · Accepted Answer

$1.9 	imes 10^{-15} \ M$

Vedclass · Answer

(C) The dissociation of $NaOH$ is: $NaOH ightarrow Na^+ + OH^-$. Since $NaOH$ is a strong electrolyte,$[OH^-] = 1.0 \ M$. Let the solubility of $Ni(OH)_2$ be $s \ M$. The dissociation is: $Ni(OH)_2 ightleftharpoons Ni^{2+} + 2OH^-$. The concentration of $Ni^{2+}$ is $s$ and the total concentration of $OH^-$ is $(2s + 1.0) \ M$. Given $K_{sp} = [Ni^{2+}][OH^-]^2 = 1.9 	imes 10^{-15}$. Substituting the values: $s(2s + 1.0)^2 = 1.9 	imes 10^{-15}$. Since $s$ is very small,we can neglect $2s$ compared to $1.0$. Thus,$s(1.0)^2 = 1.9 	imes 10^{-15}$. $s = 1.9 	imes 10^{-15} \ M$.

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:

Similar Questions

For the precipitation reaction of $Ag^{+}$ ions with $NaCl$,which of the following statements is correct?

The solubility of $CaF_2$ in a solution of $0.1 \ M \ Ca(NO_3)_2$ is.

If the solubility of $Ca_3(PO_4)_2$ in water is $x \ mol \ L^{-1}$,its solubility product in $mol^5 \ L^{-5}$ is (in $x^5$)

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

Solubility of $BaF_2$ in a solution of $Ba(NO_3)_2$ will be represented by the concentration term:

Vedclass Test Series

Exam Paper Generator

Online Exam Module