A20.0 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

The concentration of barium ions is 0.08 M

Explanation:

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

    .....(1)

Molarity of potassium carbonate = 0.200 M

Volume of solution = 20.0 mL

Putting values in equation 1, we get:

Molarity of barium nitrate = 0.400 M

Volume of solution = 30.0 mL

Putting values in equation 1, we get:

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

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 = 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 = 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:

Hence, the concentration of barium ions is 0.08 M