Question

A solution containing cacl2 is mixed with a solution of li2c2o4 to form a solution that is 2.1 × 10-5 m in calcium ion and 4.75 × 10-5 m in oxalate ion. What will happen once these solutions are mixed?

Answer 1

The question is incomplete, complete question is:

A solution containing CaCl2 is mixed with a solution of Li2C2O4 to form a solution that is
`2.1* 10^(-5) M` in calcium ion and
`4.75* 10^(-5) M` in oxalate ion. What will happen once these solutions are mixed?

`K_(sp) (CaC_2O_4) = 2.3* 10^(-9)`

Answer:

On mixing of both the solutions no precipitation will occur due to lower value of an ionic product of calcium oxalate from its solubility product.

Step-by-step explanation:

Molar concentration of calcium ions =
`[Ca^(2+)]=2.1* 10^(-5) M`

Molar concentration of oxalate ions =
`[C_2O_4^(2-)]=4.75* 10^(-5) M`

ionic product of calcium oxalate in solution :

`K_i=2.1* 10^(-5) M* 4.75* 10^(-5)M=9.975* 10^(-10)`

Solubility product of calcium oxalate =
`K_(sp)=2.3* 10^(-9)`

Generally precipitation occurs when ionic product of substance in solution exceeds its solubility product.

`K_i>K_(sp)` (precipitation occurs)

`K_i<K_(sp)` (non precipitation occurs)

`K_i<K_(sp)`

`9.975* 10^(-10)<2.3* 10^(-9)`

On mixing of both the solutions no precipitation will occur due to lower value of an ionic product of calcium oxalate from its solubility product.