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 =
Moles of silver nitrate in 10 mL of 1 M silver nitrate.
Volume of the silver nitrate solution = 10 mL = 0.010 L
Moles of silver nitrate =
Moles of sodium chromate in 25 mL of 0.1 M silver nitrate.
Volume of the silver chromate solution = 25 mL = 0.025 L
Moles of sodium nitrate =
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 :
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:
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:
Concentration of nitrate ions = 0.2857 mol/L