Question

Silver nitrate, AgNO3, can be used to test for the presence of chloride ions in solution, because it readily forms a precipitate of AgCl. What volume of 1.5 M AgNO3 will be required to react with 30.0 mL of a 0.45 M HCl solution?

Answer 1

Answer:

$\boxed{\text{9.0 mL}}$

Explanation:

1. Write the balanced chemical equation.

AgNO₃ + HCl ⟶ AgCl + HNO₃

2. Calculate the moles of HCl

$\text{Moles of HCl} =\text{30.0 mL HCl} \times \dfrac{\text{0.45 mmol HCl}}{\text{1 mL HCl}} = \text{13.5 mmol HCl}$

3. Calculate the moles of AgNO₃.

$\text{Moles of AgNO}_{3} =\text{13.5 mmol HCl} \times \dfrac{\text{1 mmol AgNO}_{3}}{\text{1 mmol HCl}} = \text{13.5 mmol AgNO}_{3}$

4. Calculate the volume of AgNO₃

$c = \text{13.5 mmol AgNO}_{3} \times \dfrac{\text{1 mL AgNO}_{3}}{\text{1.5 mmol AgNO}_{3}} = \text{9.0 mL AgNO}_{3}$

The titration will require $\boxed{\textbf{9.0 mL}}$ of AgNO₃.