Question

A 20.0 mL sample of 0.200 M K₂CO₃ solution is added to 30.0 mL of 0.400 M Ba(NO₃)₂ solution. Barium carbonate precipitates.
The concentration of barium ion, Ba²⁺, in solution after the reaction is ______.

Answer 1

Answer: The concentration of barium ions is 0.08 M

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$     .....(1)

• For potassium carbonate:

Molarity of potassium carbonate = 0.200 M

Volume of solution = 20.0 mL

Putting values in equation 1, we get:

$0.200M=\frac{\text{Moles of potassium carbonate}\times 1000}{20.0}\\\\\text{Moles of potassium carbonate}=\frac{(0.200\times 20.0)}{1000}=0.004mol$

• For barium nitrate:

Molarity of barium nitrate = 0.400 M

Volume of solution = 30.0 mL

Putting values in equation 1, we get:

$0.400M=\frac{\text{Moles of barium nitrate}\times 1000}{30.0}\\\\\text{Moles of barium nitrate}=\frac{(0.400\times 30.0)}{1000}=0.012mol$

The chemical equation for the reaction of potassium carbonate and barium nitrate follows:

$K_2CO_3(aq.)+Ba(NO_3)_2(aq.)\rightarrow BaCO_3(s)+2KNO_3(aq.)$

By Stoichiometry of the reaction:

1 mole of potassium carbonate reacts with 1 mole of barium nitrate

So, 0.004 moles of potassium carbonate will react with = $\frac{1}{1}\times 0.004=0.004mol$ of barium nitrate

As, given amount of barium nitrate is more than the required amount. So, it is considered as an excess reagent.

Thus, potassium carbonate is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of potassium carbonate produces 1 mole of barium carbonate

So, 0.004 moles of potassium carbonate will produce = $\frac{1}{1}\times 0.004=0.004mol$ of barium carbonate

We know that:

1 mole of barium carbonate produces 1 mole of barium ions and 1 mole of carbonate ions

Moles of barium ions = 0.004 moles

Volume of solution = (20 + 30) mL = 50 mL

Putting values in equation 1, we get:

$\text{Concentration of }Ba^{2+}\text{ ions}=\frac{0.004\times 1000}{50}\\\\\text{Concentration of }Ba^{2+}\text{ ions}=0.08M$

Hence, the concentration of barium ions is 0.08 M