Chapter 9, Problem 50A

Interpretation Introduction

Interpretation: The amount of silver nitrate required to completely precipitate chloride ions and the mass of silver chloride obtained is to be calculated.

Concept introduction: The hard solid mass obtained during chemical reaction which is insoluble in water is known as precipitate.

Answer to Problem 50A

The amount of silver nitrate required to completely precipitate chloride ions and the mass of silver chloride obtained is $0.52\text{g}$ and $0.44\text{g}$ , respectively.

Explanation of Solution

The total mass of the sample is $1.054\text{g}$ . The mass percentage of chloride ions in sample is $10.3\%$ . The amount of chloride ions present in the sample is calculated by the formula,

  $m_{{\text{Cl}}^{-}}=m_{\text{S}}\times \frac{10.3}{100}$

Where,

• $m_{\text{S}}$ is the mass of sample.

Substitute the value of mass of sample in the above formula.

  $\begin{array}{c}{m}_{{\text{Cl}}^{-}}=1.054\text{g}\times \frac{10.3}{100}\\ =0.108562\text{g}\end{array}$

The number of moles of chloride ions is calculated as,

  $n_{{\text{Cl}}^{-}}=\frac{m_{{\text{Cl}}^{-}}}{M_{{\text{Cl}}^{-}}}$

Where,

• $m_{{\text{Cl}}^{-}}$ is the mass of chloride ions.
• $M_{{\text{Cl}}^{-}}$ is the molar mass of chloride ions $\left(35.45\text{g/mol}\right)$ .

Substitute the value of mass and molar mass of chloride ions in the above formula.

  $\begin{array}{c}{n}_{{\text{Cl}}^{-}}=\frac{0.108562\text{g}}{35.45\text{g/mol}}\\ =3.062\times {10}^{-3}\text{mol}\end{array}$

The chemical reaction that takes place between silver nitrate and chloride ions is,

  ${\text{AgNO}}_3+{\text{Cl}}^{-}\to \text{AgCl}+{\text{NO}}_3{}^{-}$

The molar ratio of silver nitrate and chloride ions is $1:1$ . Hence, the number of moles of silver nitrate used to completely precipitate chloride ions is $3.062\times {10}^{-3}\text{mol}$ .

The amount of silver nitrate is calculated by the formula,

  $m_{{\text{AgNO}}_3}=n_{{\text{AgNO}}_3}\times M_{{\text{AgNO}}_3}$

Where,

• $n_{{\text{AgNO}}_3}$is the number of moles of silver nitrate.
• $M_{{\text{AgNO}}_3}$ is the molar mass of silver nitrate $\left(169.87\text{g/mol}\right)$ .

Substitute the value of number of moles and molar mass of silver nitrate in the above formula.

  $\begin{array}{c}{m}_{{\text{AgNO}}_3}=3.062\times {10}^{-3}\text{mol}\times 169.87\text{g/mol}\\ =0.52\text{g}\end{array}$

The molar ratio of silver chloride and chloride ions is $1:1$ . Hence, the number of moles of silver chlorideproduced by precipitation of chloride ions is $3.062\times {10}^{-3}\text{mol}$ .

The amount of silver chloride is calculated by the formula,

  $m_{\text{AgCl}}=n_{\text{AgCl}}\times M_{\text{AgCl}}$

Where,

• $n_{\text{AgCl}}$ is the number of moles of silver chloride.
• $M_{\text{AgCl}}$ is the molar mass of silver chloride $\left(143.32\text{g/mol}\right)$ .

Substitute the value of number of moles and molar mass of silver chloride in the above formula.

  $\begin{array}{c}{m}_{\text{AgCl}}=3.062\times {10}^{-3}\text{mol}\times 143.32\text{g/mol}\\ =0.44\text{g}\end{array}$

Therefore, the amount of silver nitrate required to completely precipitate chloride ions and the mass of silver chloride obtained is $0.52\text{g}$ and $0.44\text{g}$ , respectively.

Conclusion

The amount of silver nitrate required to completely precipitate chloride ions and the mass of silver chloride obtained is $0.52\text{g}$ and $0.44\text{g}$ , respectively.