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

To find the volume of 1.5 M AgNO3 required to react with 30.0 mL of a 0.45 M HCl solution, we can use the balanced equation. First, calculate the moles of HCl using volume × concentration. Then, calculate the volume of 1.5 M AgNO3 using the formula: moles ÷ concentration.

Step-by-step explanation:

To find the volume of 1.5 M AgNO3 required to react with 30.0 mL of a 0.45 M HCl solution, we can use the balanced equation:

AgNO3 (aq) + HCl (aq) → AgCl (s) + HNO3 (aq)

First, we need to calculate the moles of HCl:

Moles of HCl = volume (L) × concentration (M)

Moles of HCl = 0.030 L × 0.45 M = 0.0135 mol

According to the balanced equation, 1 mole of HCl reacts with 1 mole of AgNO3. Therefore, we need 0.0135 mol of AgNO3.

Next, we can calculate the volume of 1.5 M AgNO3 using the formula:

Volume (L) = moles ÷ concentration (M)

Volume (L) = 0.0135 mol ÷ 1.5 M = 0.009 L = 9.0 mL

Answer 2

`\boxed{\text{9.0 mL}}`

Step-by-step 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} * \frac{\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} * \frac{\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) * \frac{\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.