Question

How many grams of silver chromate will precipitate when $150 {\text{mL}}$ of $0.5 {\text{M}}$ silver nitrate is added to $100 {\text{mL}}$ of $0.4 {\text{M}}$ potassium chromate?

Answer 1

First, we will write the above-mentioned chemical reaction and balance it. Then we will find the equivalent mass of silver and chromate ion to get the mass of silver chromate precipitated in the solution. We shall find the moles of the reactants and the limiting reagent. Then we shall find the moles of silver chromate formed and thus its weight.

Formula Used
We would require the formula for molarity to solve this question
${\text{molarity}} = \dfrac{{{n_s}}}{{{v_s}}}$
Where, ${n_s}$ is the number of moles of solute
${v_s}$ is the volume of solvent in litres.

Complete step by step solution:
Let us start with writing the balanced chemical reaction
$2A{g^ + } + CrO_4^{2 - } \to A{g_2}Cr{O_4}(s) \downarrow$
We can observe that silver chromate will deposit as a precipitate of brick red colour.
According to the question, the following information is provided to us
Molarity of silver nitrate solution $= 0.5{\text{ M}}$
Volume of silver nitrate solution $= 150{\text{ mL}} = 0.15{\text{ }}{\text{L}}$
Molarity of potassium chromate solution $= 0.4{\text{ M}}$
Volume of silver potassium chromate $= 100{\text{ mL}} = 0.1{\text{ L}}$
We know that molarity tells us the number of moles of solute per unit volume of solvent (in litres). Now, we will use the above formula to get the number of moles of silver nitrate and potassium chromate
${\text{molarity}} = \dfrac{{{n_s}}}{{{v_s}}}$
We can rewrite it as
${n_s} = {\text{molarity}} \times {v_s}$
Upon substituting values of silver nitrate, we get
${n_s} = 0.5{\text{M}} \times 0.15{\text{L}}$
Upon solving, we get ${n_s} = 0.075$ moles of silver nitrate
Upon substituting values of potassium chromate, we get
${n_s} = 0.4{\text{M}} \times 0.1{\text{ L}}$
Upon solving, we get ${n_s} = 0.04$ moles of potassium chromate
Clearly, silver nitrate is the limiting reagent, as there is a stoichiometric excess of chromate ion.
Also, we know the molar mass of $A{g_2}Cr{O_4} = 331.73{\text{ g mo}}{{\text{l}}^{{\text{ - 1}}}}$
Look at the stoichiometric coefficients on the reactant and product side. They are in the ratio $2:1$ . So, we need to divide the number of moles by $2$ to get the final answer.
So, the mass of silver chromate precipitated is given by
$\dfrac{1}{2} \times 0.0750 {\text{mol}} \times 331.73{\text{ g mo}}{{\text{l}}^{{\text{ - 1}}}} = 12.4 {\text{g}}$
Hence, $12.4 g$ of silver chromate will precipitate when $150 {\text{mL}}$ of $0.5 {\text{M}}$ silver nitrate is added to $100 {\text{mL}}$ of $0.4 {\text{M}}$ potassium chromate.

Note:
Precipitation reactions are generally double displacement reactions that involve the production of a residue of a solid form called the precipitate. These reactions also occur when two or more solutions are combined with different salts, resulting in the formation of insoluble salts from which the solution is precipitated.