# There are 2.8 * 10^-2 mol of Ba(2+) ions in 1 dm^3 of a Barium salt solution. 20 cm^3 of (NH4)2C2O4 solution has been added to 20 cm^3 of above solution. In this time this solution has been faintly...

There are 2.8 * 10^-2 mol of Ba(2+) ions in 1 dm^3 of a Barium salt solution. 20 cm^3 of (NH4)2C2O4 solution has been added to 20 cm^3 of above solution. In this time this solution has been faintly saturated by BaC2O4 salt. In the related temperature if Ksp of BaC2O4 is 1.4 * 10^-9 mol^2dm^-6,**find the weight of (NH4)2C2O4 in 1 dm^3 of ammonium oxalate solution.**

(N=14, C=12, O=16 , H=1)

**Please help me to slove this.**

### 1 Answer | Add Yours

We have mixed 20ml of each solution to get the faintly saturated.

When the each 20ml two solutions mixed the concentration of each solution will get halved.

Since the solution is faintly saturated there will be no precipitates.

`[Ba^(2+)] = (2.8xx10^(-2))/2 = 1.4xx10^(-2)`

Since there is a faint saturation we can use the `K_(sp)` equation.

`Ba^(2+)+C_2O_4^(2-) rarr BaC_2O_4`

`K_(sp) = [Ba^(2+)][C_2O_4^(2-)]`

`1.4xx10^(-9) = 1.4xx10^(-4)[C_2O_4^(2-)]`

`[C_2O_4^(2-)] = 1xx10^(-5)`

`(NH_4)_2C_2O_4 rarr 2NH_4^++C_2O_4^(2-)`

Mole ratio

`(NH_4)_2C_2O_4:C_2O_4^(2-) = 1:1`

`[C_2O_4^(2-)] = [(NH_4)_2C_2O_4] = 1xx10^(-5)`

molar mass of `(NH_4)_2C_2O_4 = 124g/(mol)`

* Weight of *`1dm^3` of `(NH_4)_2C_2O_4`

`= 124xx1xx10^(-5) `

`= 0.00124g`

*So weight of 1dm^3 of ammonium oxalate solution is 0.00124g.*

**Sources:**