Question

A 100-milliliter solution containing $AgN{O}_3$ was treated with excess $NaCl$ to completely precipitate the silver as $AgCl$. If 5.7 g $AgCl$ was obtained, what was the concentration of $A{g}^{+}$ in the original solution?

Answer 1

$AgN{O}_3+NaOH\to AgCl+NaN{O}_3$

molecular mass of $AgCl$ = 143.5 g

moles of $AgCl$ produced = $\frac{5.7}{143.5}$ = $0.04$

1 mole of $AgCl$ is produced by 1 mole of $AgN{O}_3$

so, the moles of $A{g}^{+}$ ion = $0.04$ moles

concentration of $A{g}^{+}$ ion = $\frac{moles}{volumeofsolution\left(L\right)}$ = $\frac{0.04\ast 1000}{100}$

= $0.4M$