Equal volumes of 0.02 M CaCl_2 and 0.00004 | vedclass

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

Question

(B) When equal volumes are mixed,the total volume doubles,so the concentration of each species is halved. $[Ca^{2+}] = \frac{0.02}{2} = 0.01 \ M = 10^

Vedclass · Accepted Answer

No,precipitation will not occur.

Vedclass · Answer

(B) When equal volumes are mixed,the total volume doubles,so the concentration of each species is halved. $[Ca^{2+}] = \frac{0.02}{2} = 0.01 \ M = 10^{-2} \ M$. $[SO_4^{2-}] = \frac{0.00004}{2} = 0.00002 \ M = 2 	imes 10^{-5} \ M$. The ionic product $Q_{sp} = [Ca^{2+}][SO_4^{2-}] = (10^{-2}) 	imes (2 	imes 10^{-5}) = 2 	imes 10^{-7}$. Since $Q_{sp} (2 	imes 10^{-7}) < K_{sp} (2.4 	imes 10^{-5})$,the solution is unsaturated and no precipitation of $CaSO_4$ will occur.

Equal volumes of $0.02 \ M$ $CaCl_2$ and $0.00004 \ M$ $Na_2SO_4$ solutions are mixed. Will a precipitation of $CaSO_4$ occur? $(K_{sp} = 2.4 \times 10^{-5})$

Similar Questions

The $K_{sp}$ of $M(OH)_2$ is $3.2 \times 10^{-11}$. The $pH$ of its saturated solution in water is:

If the concentration of $CrO_4^{2-}$ ions in a saturated solution of silver chromate is $2 \times 10^{-4} \ M$,the solubility product of silver chromate will be:

The solubility product of $BaSO_4$ is $1 \times 10^{-10}$ at $298 \ K$. The solubility of $BaSO_4$ in $0.1 \ M \ K_2SO_4 \ (aq)$ solution is $......... \times 10^{-9} \ g \ L^{-1}$ (nearest integer). Given: Molar mass of $BaSO_4$ is $233 \ g \ mol^{-1}$.

At $30\,^{\circ}C,$ the solubility of $Ag_2CO_3$ $(K_{sp} = 8 \times 10^{-12})$ would be greatest in one litre of

The solubility of $CuBr$ is $2 \times 10^{-4} \ mol/L$ at $25 \ ^\circ C$. The $K_{sp}$ value for $CuBr$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module