Question

A mixture of solid srso4 and Baso4 is shaking with watern until solution is estblished ,calculate sr2 +,Ba2 +,so42-. ksp=Srso4=7.6*10-7 molarsquare,and ksp=Baso4 =1.5*10-7

Answer 1

Answer $[Sr^{2+}]=0.871\times 10^(-3) mol/L,[Ba^{2+}]=0.387\times 10^{-3} mol/L,[SO_4^{2-}]=1.258\times 10^{-3}mol/L$

Solution:

1) $SrSO_4\rightleftharpoons Sr^{2+}+SO_4^{2-}$

              $S_1$           $S_1$

$K_{sp} of SrSO_4=S_1\times S_1=S_1^2=7.6\times 10^{-7} (mol/L)^2$

$S_1=0.871\times 10^{-3} mol/L$

The concentration of strontium ion:

$[Sr^{2+}]=0.871\times 10^{-3} mol/L$

2) $BaSO_4\rightleftharpoons Ba^{2+}+SO_4^{2-}$

              $S_2$           $S_2$

$K_{sp} of BaSO_4=S_2\times S_2=S_2^2=1.5\times 10^{-7}(mol/L)^2$

$S_2=.0387\times 10^{-3} mol/L$

The concentration of barium ion:

$[Ba^{2+}]=0.387\times 10^{-3} mol/L$

3) Concentration of$[SO_4^{2-}]$:

Since $SO_4^{2-}$ ion is coming from both the salts then  $[SO_4^{2-}]$ will be:

$=S_1+S_2=0.871\times 10^(-3) mol/L+0.387\times 10^{-3} mol/L=1.258\times 10^{-3}mol/L$

Answer 2

Answer: The concentrations of

$Sr^{2+}=8.7\times 10^{-4}M$

$Ba^{2+}=3.8\times 10^{-4}M$

$SO_4^{2-}=12.5\times 10^{-4}M$

Explanation:

To calculate the concentration of the ions, we need to write the equations for the ionization of these salts.

1. Concentration of $Sr^{2+]$

 $SrSO_4\rightleftharpoons Sr^{2+}+SO_4^{2-}$

                 $s_1$            $s_1$

$K_{sp}$ for this equation is written as:

$K_{sp}=[Sr^{2+}][SO_4^{2-}]=s_1\times s_1=s_1^2$

$K_{sp}$ for $SrSO_4$ = $7.6\times 10^{-7}$

Putting values, we get:

$7.6\times 10^{-7}=s_1^2\\\\s_1=8.7\times 10^{-4}$

$[Sr^{2+}]=8.7\times 10^{-4}M$

2. Concentration of $Ba^{2+]$

 $BaSO_4\rightleftharpoons Ba^{2+}+SO_4^{2-}$

                   $s_2$            $s_2$

$K_{sp}$ for this equation is written as:

$K_{sp}=[Ba^{2+}][SO_4^{2-}]=s_2\times s_2=s_2^2$

$K_{sp}$ for $BaSO_4$ = $1.5\times 10^{-7}$

Putting values, we get:

$1.5\times 10^{-7}=s_2^2\\\\s_2=3.8\times 10^{-4}$

$[Ba^{2+}]=3.8\times 10^{-4}M$

3. Concentration of $SO_4^{2-}$

As sulfate ion is present in both the salts, so total concentration of this ion will be = $s_1+s_2$

Concentration of $SO_4^{2-}=(8.7+3.8)\times 10^{-4}=12.5\times 10^{-4}M$