Chapter U4.87, Problem 8E

(a)

Interpretation Introduction

Interpretation: The number of moles of ${\text{H}}^{\text{+}}$ in 15 mL of 0.025 M HCl solution needs to be determined.

Concept Introduction: pH of a solution indicates the concentration of ${\text{H}}^{\text{+}}$ ions in the solution. The mathematical expression can be written as:

  ${\text{pH = - log [H}}^{\text{+}}\text{]}$

Similarly the pOH value is related to the concentration of OH- ions as given:

  $\begin{array}{l}{\text{pOH = - log [OH}}^{\text{-}}\text{]}\\ \text{pH + pOH = 14}\end{array}$

Answer to Problem 8E

  $\text{Moles = }0.375{\text{×10}}^{\text{-3}}\text{ moles}$

Explanation of Solution

Concentration of HCl=[ ${\text{H}}^{\text{+}}$ ] = 0.025 M

Volume = 15 mL

Calculate moles of HCl:

  $\begin{array}{l}\text{Concentration =}\frac{\text{moles}}{\text{volume}}\\ \text{0}\text{.025 M =}\frac{\text{moles}}{\text{15}{\text{.0 ×10}}^{\text{-3}}\text{ L}}\\ \text{Moles = 0}\text{.025 M ×15}{\text{.0 ×10}}^{\text{-3}}\text{ L}\\ \text{Moles = }0.375{\text{×10}}^{\text{-3}}\text{ moles}\end{array}$

(b)

Interpretation Introduction

Interpretation: The concentration of solution after addition of 25 mL of water in 15 mL of 0.025 M HCl solution needs to be determined.

Concept Introduction: pH of a solution indicates the concentration of ${\text{H}}^{\text{+}}$ ions in the solution. The mathematical expression can be written as:

  ${\text{pH = - log [H}}^{\text{+}}\text{]}$

Similarly the pOH value is related concentration of OH- ions as given:

  $\begin{array}{l}{\text{pOH = - log [OH}}^{\text{-}}\text{]}\\ \text{pH + pOH = 14}\end{array}$

Answer to Problem 8E

  ${\text{M}}_{\text{2}}\text{= 9}{\text{.37×10}}^{\text{-3}}\text{M}$

Explanation of Solution

[ ${\text{H}}^{\text{+}}$ ] = 0.025 M

Volume = 15 mL

M₁ = 0.025 M

V₁ = 15 mL

M₂ = need to calculate

V₂ = 15 + 25 = 40 mL

Calculate the new concentration:

  $\begin{array}{l}{\text{M}}_{\text{1}}{\text{×V}}_{\text{1}}{\text{ = M}}_{\text{2}}{\text{×V}}_{\text{2}}\\ \text{0}{\text{.025 M ×15 mL = M}}_{\text{2}}\text{× 40 mL}\\ {\text{M}}_{\text{2}}\text{=}\frac{\text{0}\text{.025 M ×15 mL}}{\text{40 mL}}\\ {\text{M}}_{\text{2}}\text{= 9}{\text{.37×10}}^{\text{-3}}\text{M}\end{array}$

(c)

Interpretation Introduction

Interpretation: The final pH of the solution after addition of 25 mL of water in 15 mL of 0.025 M HCl solution.

Concept Introduction: pH of a solution indicates the concentration of ${\text{H}}^{\text{+}}$ ions in the solution. The mathematical expression can be written as:

  ${\text{pH = - log [H}}^{\text{+}}\text{]}$

Similarly the pOH value is related concentration of OH- ions as given:

  $\begin{array}{l}{\text{pOH = - log [OH}}^{\text{-}}\text{]}\\ \text{pH + pOH = 14}\end{array}$

Answer to Problem 8E

  $\text{pH = }2.03$

Explanation of Solution

[ ${\text{H}}^{\text{+}}$ ] = 0.025 M

Volume = 15 mL

M₁ = 0.025 M

V₁ = 15 mL

M₂ = need to calculate

V₂ = 15 + 25 = 40 mL

Calculate the new concentration:

  $\begin{array}{l}{\text{M}}_{\text{1}}{\text{×V}}_{\text{1}}{\text{ = M}}_{\text{2}}{\text{×V}}_{\text{2}}\\ \text{0}{\text{.025 M ×15 mL = M}}_{\text{2}}\text{× 40 mL}\\ {\text{M}}_{\text{2}}\text{=}\frac{\text{0}\text{.025 M ×15 mL}}{\text{40 mL}}\\ {\text{M}}_{\text{2}}\text{= 9}{\text{.37×10}}^{\text{-3}}\text{M}\end{array}$

This is equal to the concentration of hydrogen ion.

Calculate pH:

  $\begin{array}{l}{\text{pH = - log [H}}^{\text{+}}\text{]}\\ \text{pH = - log [9}\text{.37}\times {\text{10}}^{\text{-3}}\text{M]}\\ \text{pH = }2.03\end{array}$