Answer:
[tex]\boxed{\text{9.0 mL}}[/tex]
Explanation:
1. Write the balanced chemical equation.
AgNO₃ + HCl ⟶ AgCl + HNO₃
2. Calculate the moles of HCl
[tex]\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}[/tex]
3. Calculate the moles of AgNO₃.
[tex]\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}[/tex]
4. Calculate the volume of AgNO₃
[tex]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}[/tex]
The titration will require [tex]\boxed{\textbf{9.0 mL}}[/tex] of AgNO₃.
