Question

On mixing $50.0ml$ of $0.2$ M $NaCl$ and $50.0$ mL of $0.25$ M $AgN{O}_3$, if all the $C{l}^{-}$ ions are precipitated, the molarity of $A{g}^{+}$ is:

• A. $2.5\times {10}^{-2}M$
• B. $7.5\times {10}^{-2}M$
• C. $5.2\times {10}^{-2}M$
• D. $5.7\times {10}^{-2}M$

Answer 1

The correct option is A $2.5\times {10}^{-2}M$

The reaction for the formation of the precipitate of AgCl is $A{g}_{\left(aq\right)}^{+}+C{l}_{\left(aq\right)}^{-}\to AgC{l}_{\left(s\right)}$
Initial number of millimoles of $A{g}^{+}=50\times 0.25=12.5$
Initial number of millimoles of $C{l}^{-}=50\times 0.2=10.00$
Excess millimoles of $A{g}^{+}=12.5-10=2.5$
Excess moles of $A{g}^{+}=2.5\times {10}^{-3}$
Total volume $=50.0+50.0=100$ ml
Hence, the molarity of $A{g}^{+}=\frac{\text{number of moles}}{\text{volume in ml}}\times 1000=\left(\frac{2.5\times {10}^{-3}}{100}\right)\times 1000=2.5\times {10}^{-2}M$.