# In a aqueous solution, ion concentration of CO3(2-) is 0.01 mol dm^-3 and ion concentration of SO4(2-) is 0.05 mol dm^-3. In 1.01 cm^3 of this solution, 0.01 mol of Calcium ethanoate solid has...

In a aqueous solution, ion concentration of CO3(2-) is 0.01 mol dm^-3 and ion concentration of SO4(2-) is 0.05 mol dm^-3. In 1.01 cm^3 of this solution, 0.01 mol of Calcium ethanoate solid has been added bit by bit.In the given temperature,If Ksp of CaSO4 is 9 * 10^-6 mol^2 dm^-6 and Ksp of CaCO3 is 1.21 * 10^-10 mol^2dm^-6, **what will be the mass of the precipitation which will be precipitated?**

**Please help me on this**.

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### 1 Answer

Ionic concentration of `CO_3^(2-)` is 0.01 mol dm^-3 and ion concentration of `SO_4^(2-)` is 0.05 mol dm^-3.

In 1.01 cm^3 of this solution, 0.01 mol of Calcium ethanoate solid has been added bit by bit. That makes the ionic concentration of `Ca^(2+)` = 0.01*1000/1.01=10/1.01 M in that part of the solution.

Therefore, maximum allowable concentration of `CO_3^(2-)` in 1.01 portion is `K_(sp)/(Ca^(2+))=(1.21*10^(-10)*1.01)/10`

`=1.222*10^(-11) M`

Moles of `CO_3^(2-)` deposited = moles taken- moles remaining in soln

`= ((0.01-1.222*10^(-11))*1.01)/1000`

Mass of `CaCO_3` deposited

= moles of `CO_3^(2-) ` * molar mass of `CaCO_3`

`=(100*0.01*1.01)/1000=1.01*10^(-3) g`

Similarly,

Maximum allowable concentration of `SO_4^(2-)` in 1.01 cm^3 portion is `K_(sp)/(Ca^(2+))=(9*10^(-6)*1.01)/10`

`=9.09*10^(-7) M`

Moles of `SO_4^(2-) ` deposited = moles taken - moles remaining in soln

`= ((0.05-9.09*10^(-7))*1.01)/1000`

Mass of `CaSO_4 ` deposited

= moles of `SO_4^(2-)` * molar mass of `CaSO_4`

`=(136*0.05*1.01)/1000=6.87*10^(-3) g`

Therefore, the total mass of the precipitate that is formed is `(6.87+1.01)*10^(-3) g`

=`7.88*10^(-3) g`

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