Question

The ksp of manganese(ii) carbonate, mnco3, is 2.42 × 10-11. calculate the solubility of this compound in g/l.

Answer 1

To calculate the solubility of MnCO3 in g/L, use the solubility product constant (Ksp). The solubility of MnCO3 comes out to be approximately 1.56 × 10-6 g/L.

Step-by-step explanation:

To calculate the solubility of manganese(II) carbonate (MnCO3) in g/L, we need to use the solubility product constant (Ksp). The Ksp value for MnCO3 is 2.42 × 10-11.

Let's assume the solubility of MnCO3 is 'x' g/L.

When MnCO3 dissolves in water, it dissociates into Mn2+ and CO32- ions.

According to the balanced equation for the solubility equilibrium, the concentration of both ions will be 'x'.

Therefore, we can write the solubility product expression for MnCO3 as: Ksp = [Mn2+][CO32-].

Substituting the values, we get: 2.42 × 10-11 = x × x.

Solving this equation, we find the solubility of MnCO3 to be approximately 1.56 × 10-6 g/L.

Answer 2

Answer : The solubility of this compound in g/L is
`565.414* 10^(-6)g/L`.

Solution : Given,

`K_(sp)=2.42* 10^(-11)`

Molar mass of
`MnCO_3` = 114.945g/mole

The balanced equilibrium reaction is,

`MnCO_3\rightleftharpoons Mn^(2+)+CO^(2-)_3`

At equilibrium s s

The expression for solubility constant is,

`K_(sp)=[Mn^(2+)][CO^(2-)_3]`

Now put the given values in this expression, we get

`2.42* 10^(-11)=(s)(s)\\2.42* 10^(-11)=s^2\\s=0.4919* 10^(-5)=4.919* 10^(-6)moles/L`

The value of 's' is the molar concentration of manganese ion and carbonate ion.

Now we have to calculate the solubility in terms of g/L multiplying by the Molar mass of the given compound.

`s=4.919* 10^(-6)moles/L* 114.945g/mole=565.414* 10^(-6)g/L`

Therefore, the solubility of this compound in g/L is
`565.414* 10^(-6)g/L`.