Question

10mL of HCl solution produced 0.1435g of A gcl when treated with excesss of silver nitrate solution. What is the molarity of the acid solution?(at. mass Ag=108)

Answer 1

Reaction between HCl and silver nitrate:
$$\begin{gathered} HCl+AgN{O}_3\to AgCl+HN{O}_3 \\ 36.5g143.5g \end{gathered}$$

143.5 g AgNO₃ is produced from 36.5 g HCl
0.1435 g of AgNO₃ is produced from (36.5/143.5) x 0.1435 = 0.0365 g HCl

Now, given volume of HCl = 10 ml
Therefore,
10 ml HCl contains 0.0365 g HCl
$$\begin{gathered} \mathrm{Molarity}\mathrm{of}solution\; =\frac{0.0365}{36.5}x\frac{1000}{10} \\ =0.1M \end{gathered}$$