Question

10mL of 1M silver nitrate is combined with 25mL of 0.1M sodium chromate. What is the concentration of sodium ion after the two solutions are combined? This set up will be used for the next two questions as well, so you should also calculate the combined concentrations of the other ions.

Answer 1

Answer:

Concentration of sodium ions in final solution is 0.1428 mol/L.

Concentration of nitrate ions in final solution is 0.2857 mol/L.

Explanation:

Concentration =$\frac{\text{Moles of compound}}{\text{Volume of the solution (L)}}$

$Na_2CrO_4(aq)+2AgNO_3(aq)\rightarrow Ag_2CrO_4(s)+2NaNO_3(aq)$

Moles of silver nitrate in 10 mL of 1 M silver nitrate.

Volume of the  silver nitrate solution = 10 mL = 0.010 L

$1 M =\frac{\text{Moles of}Ag_2CrO_4}{\text{Volume of the solution in L}}$

Moles of silver nitrate =$1 M\times 0.010 L=0.010 mol$

Moles of sodium chromate in 25 mL of 0.1 M silver nitrate.

Volume of the  silver chromate solution = 25 mL = 0.025 L

$0.1 M =\frac{\text{Moles of}Na_2CrO_4}{\text{Volume of the solution in L}}$

Moles of sodium nitrate =$0.1 M\times 0.025 L=0.0025 mol$

According to reaction, 1 mole of sodium chromate reacts with 2 mole of silver nitrate.

Then , 0.0025 mol of sodium chromate will react with :

$\frac{1}{2}\times 0.0025 mol=0.00125 mol$ of silver nitrate.

1 mol of sodium chromate gives 2 mol of sodium ions and 1 mol of chromate ions.

Then 0.0025 mol of sodium chromate will give 0.0050 mol of sodium ions.

Volume of the solution after mixing =

10 mL + 25 mL = 35 mL =0.035 L

Concentration of sodium ions in final solution:

$[Na^+]=\frac{0.0050 mol}{0.035 mL}=0.1428 mol/L$

Concentration of sodium ions = 0.1428 mol/L

1 mole of silver nitrate gives 1 mol of silver ion and 1 mole nitrate ion

Then 0.010 moles of silver nitrate will give 0.010 moles of nitrate ions.

Concentration of nitrate ion in the final solution:

$[NO_3^{-}]=\frac{0.010 mol}{0.035 L}=0.2857 mol/L$

Concentration of nitrate ions = 0.2857 mol/L