Question

Does anyone know how to do this? the solubility of barium carbonate, BaCO3, is 0.0100 g/L. Its molar mass is 197.3 g/mol. What is the Ksp of barium carbonate?

Answer 1

The value of ${{\text{K}}_{{\text{sp}}}}$ for barium carbonate is $\boxed{{\text{2}}{{.6 \times 1}}{{\text{0}}^{{\text{ - 9}}}}}$.

Further explanation:

Solubility product constant:

The equilibrium constant between the compound and its ion, when dissolved in solution, is known as solubility product constant. It is denoted by ${{\text{K}}_{{\text{sp}}}}$. The solubility product constant is used to calculate the product of the concentration of ions at equilibrium.

Higher the solubility product constant more will be the solubility of the compound.

The general reaction is as follows:

${\text{AB}}\left({aq}\right)\to{{\text{A}}^+}\left({aq}\right)+{{\text{B}}^-}\left({aq}\right)$

The expression to calculate the solubility product for the general reaction is as follows:

$\boxed{{{\text{K}}_{{\text{sp}}}}=\left[{{{\text{A}}^+}}\right]\left[{{{\text{B}}^-}}\right]}$

Here,

${{\text{K}}_{{\text{sp}}}}$ is the solubility product constant.

$\left[ {{{\text{A}}^ + }} \right]$ is the concentration of ${{\text{A}}^ + }$ions.

$\left[ {{{\text{B}}^ - }} \right]$is the concentration of ${{\text{B}}^ - }$ions.

The dissociation of ${\text{BaC}}{{\text{O}}_{\text{3}}}$ occurs as follows:

${\text{BaC}}{{\text{O}}_3}\to{\text{B}}{{\text{a}}^{2+}}+{\text{CO}}_3^{2-}$

The given solubility of ${\text{BaC}}{{\text{O}}_{\text{3}}}$ is 0.0100 g/L. Firstly, it is to be converted in mol/L. So the solubility of ${\text{BaC}}{{\text{O}}_{\text{3}}}$ is calculated as follows:

${\text{Solubility of BaC}}{{\text{O}}_{\text{3}}}\left({{\text{mol/L}}}\right)= \frac{{{\text{Solubility of BaC}}{{\text{O}}_{\text{3}}}\left({{\text{g/L}}}\right)}}{{{\text{Molar mass of BaC}}{{\text{O}}_{\text{3}}}\left({{\text{g/mol}}}\right)}}$      …… (1)

The solubility of ${\text{BaC}}{{\text{O}}_{\text{3}}}$ is 0.0100 g/L.

The molar mass of ${\text{BaC}}{{\text{O}}_{\text{3}}}$ is 197.3 g/mol.

Substitute these values in equation (1).

$\begin{aligned}{\text{Solubility of BaC}}{{\text{O}}_{\text{3}}}&=\left( {\frac{{{\text{0}}{\text{.0100 g}}}}{{{\text{1 L}}}}}\right)\left({\frac{{{\text{1 mol}}}}{{{\text{197}}{\text{.3 g}}}}} \right)\\&=0.00005068\;{\text{mol/L}}\\\end{aligned}$

It is evident from the chemical equation, one mole of ${\text{BaC}}{{\text{O}}_{\text{3}}}$ dissociates to form one mole of ${\text{B}}{{\text{a}}^{2+}}$ and one mole of ${\text{CO}}_3^{2-}$. So the solubility of both ${\text{B}}{{\text{a}}^{2+}}$ and ${\text{CO}}_3^{2-}$is 0.00005068 mol/L.

The formula to calculate the solubility product of  is as follows:

${{\text{K}}_{{\text{sp}}}}=\left[{{\text{B}}{{\text{a}}^{{\text{2+}}}}}\right]\left[{{\text{CO}}{{_3^2}^-}}\right]$                                      …… (2)

Substitute 0.00005068 mol/L for$\left[{{\text{B}}{{\text{a}}^{{\text{2+}}}}}\right]$ and 0.00005068 mol/L for $\left[{{\text{CO}}_3^{2-}}\right]$in equation (2).

$\begin{aligned}{{\text{K}}_{{\text{sp}}}}&=\left({{\text{0}}{\text{.00005068}}} \right)\left({{\text{0}}{\text{.00005068}}}\right)\\&=2.6\times{10^{-9}}\\\end{aligned}$

Therefore, the value of ${{\mathbf{K}}_{{\mathbf{sp}}$ for ${\mathbf{BaC}}{{\mathbf{O}}_{\mathbf{3}$ is ${\mathbf{2}}{\mathbf{.6 \times 1}}{{\mathbf{0}}^{-9}}$.

Learn more:

1. Sort the solubility of gas will increase or decrease: brainly.com/question/2802008.

2. What is the pressure of the gas?: brainly.com/question/6340739.

Answer details:

Grade: School School

Subject: Chemistry

Chapter: Chemical equilibrium

Keywords: solubility, Ba2+, CO32-, BaCO3, Ksp, solubility product, molar mass, 197.3 g/mol, mol/L, g/L, 0.0100 g/L.

Answer 2

It's simple, just follow my steps.

1º - in 1 L we have $0.0100~g$ of $BaCO_3$

2º - let's find the number of moles.

$\eta=\frac{m}{MM}$

$\eta=\frac{0.0100}{197.3}$

$\boxed{\boxed{\eta=5.07\times10^{-5}~mol}}$

3º - The concentration will be

$C=5.07\times10^{-5}~mol/L$

But we have this reaction

$BaCO_3\rightleftharpoons Ba^{2+}+CO_3^{2-}$

This concentration will be the concentration of $Ba^{2+}~~and~~CO_3^{2-}$

$K_{sp}=\frac{[Ba^{2+}][CO_3^{2-}]}{[BaCO_3]}$

considering $[BaCO_3]=1~mol/L$

$K_{sp}=[Ba^{2+}][CO_3^{2-}]$

and

$[Ba^{2+}]=[CO_3^{2-}]=5.07\times10^{-5}~mol/L$

We can replace it

$K_{sp}=(5.07\times10^{-5})*(5.07\times10^{-5})$

$K_{sp}\approx25.70\times10^{-10}$

Therefore the $K_{sp}$ is:

$\boxed{\boxed{\boxed{K_{sp}\approx2.57\times10^{-11}}}}$