Chemistry, asked by yashpk3351, 1 year ago

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

Answers

Answered by IlaMends
11

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

Answered by RomeliaThurston
4

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

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