A solution is saturated with SrCO_3 and SrF_2 | vedclass

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

Question

(D) For $SrCO_3$, the solubility product expression is $K_{sp} = [Sr^{2+}][CO_3^{2-}]$. Given $[CO_3^{2-}] = 1.2 	imes 10^{-3} \ M$ and $K_{sp}(SrCO_

Vedclass · Accepted Answer

$3.7 	imes 10^{-2} \ M$

Vedclass · Answer

(D) For $SrCO_3$, the solubility product expression is $K_{sp} = [Sr^{2+}][CO_3^{2-}]$. Given $[CO_3^{2-}] = 1.2 	imes 10^{-3} \ M$ and $K_{sp}(SrCO_3) = 7.0 	imes 10^{-10}$, we calculate the concentration of $Sr^{2+}$: $[Sr^{2+}] = \frac{7.0 	imes 10^{-10}}{1.2 	imes 10^{-3}} = 5.83 	imes 10^{-7} \ M$. For $SrF_2$, the solubility product expression is $K_{sp} = [Sr^{2+}][F^{-}]^2$. Substituting the values: $7.9 	imes 10^{-10} = (5.83 	imes 10^{-7}) 	imes [F^{-}]^2$. $[F^{-}]^2 = \frac{7.9 	imes 10^{-10}}{5.83 	imes 10^{-7}} = 1.355 	imes 10^{-3}$. $[F^{-}] = \sqrt{1.355 	imes 10^{-3}} \approx 3.68 	imes 10^{-2} \ M \approx 3.7 	imes 10^{-2} \ M$.

$A$ 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: $K_{sp}(SrCO_3) = 7.0 \times 10^{-10}$, $K_{sp}(SrF_2) = 7.9 \times 10^{-10}$

Similar Questions

$A$ precipitate is formed when:

Which of the following formulas can be used to calculate the solubility of $Ag_2CO_3$?

At $25 \, ^\circ C$,the solubility product $(K_{sp})$ of $Mg(OH)_2$ is $1.0 \times 10^{-11}$. What is the minimum $pH$ required to precipitate $Mg(OH)_2$ from a $0.001 \, M \, Mg^{2+}$ solution?

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

If the concentration of $Cr_2O_7^{2-}$ in a saturated solution of $Ag_2Cr_2O_7$ is $6.5 \times 10^{-5} \ M$ at a constant temperature,calculate the value of $K_{sp}$ for $Ag_2Cr_2O_7$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module

$A$ 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: $K_{sp}(SrCO_3) = 7.0 \times 10^{-10}$, $K_{sp}(SrF_2) = 7.9 \times 10^{-10}$