Chapter 4, Problem 109GQ

1. (a) What is the pH of a 0.105 M HCl solution?
2. (b) What is the hydronium ion concentration in a solution with a pH of 2.56? Is the solution acidic or basic?
3. (c) A solution has a pH of 9.67. What is the hydronium ion concentration in the solution? Is the solution acidic or basic?
4. (d) A 10.0-mL sample of 2.56 M HCl is diluted with water to 250. mL What is the pH of the dilute solution?

Interpretation Introduction

Interpretation:

$\text{pH}$ of $\text{0}\text{.105 M HCl}$ solution has to be determined.

Concept introduction:

• Strong acids dissociates completely into ions in solution but weak acids do not.
• pH of a solution is the negative of the base $\text{-10}$ logarithm of the hydronium ion concentration.

  $\text{pH}\text{=}\text{-log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$

• Concentration of hydronium ion $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{=}{\text{10}}^{\text{-pH}}$
• For an acidic solution $\text{pH}\text{<7}$ and for a basic solution $\text{pH>}\text{7}$.
• $Amountofsubs\tan ce=Concnetrationofthesubs\tan ce\times Volume$

Answer to Problem 109GQ

$pH$ of $0.105MHCl$ solution is $\mathrm{0.979.}$

Explanation of Solution

$\text{pH}$ Of a solution is the negative of the base $\text{-10}$ logarithm of the hydronium ion concentration.

  $\text{pH}\text{=}{\text{-log}}_{10}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$

It possible to substitute the value of $H^{+}$ instead of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$

$HCl$ is a strong acid.  So the concentration of $H^{+}andHCl$ will be equal.

  $\begin{array}{l}[H^{+}]=[HCl]\\ [H^{+}]=0.015M\\ \text{pH}\text{=}-{\text{log}}_{10}\left[{\text{H}}^{\text{+}}\right]\\ =-\log 0.105\\ =0.979\end{array}$

Interpretation Introduction

Interpretation:

Hydronium ion concentration in a solution with a $pHof2.56$ has to be determined and the solution is whether acidic or basic has to be ascertained.

Concept introduction:

• Strong acids dissociates completely into ions in solution but weak acids do not.
• pH of a solution is the negative of the base $\text{-10}$ logarithm of the hydronium ion concentration.

  $\text{pH}\text{=}\text{-log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$

• Concentration of hydronium ion $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{=}{\text{10}}^{\text{-pH}}$
• For an acidic solution $\text{pH}\text{<7}$ and for a basic solution $\text{pH>}\text{7}$.

Answer to Problem 109GQ

$\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$=$\text{0}\text{.0028}\text{M}$.  The solution is acidic.

Explanation of Solution

Concentration of hydronium ion can be determined using the formula,

  Concentration of hydronium ion $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{=}{\text{10}}^{\text{-pH}}$

The pH of the given solution is $2.56$.  Substituting this value in the above equation,

Then concentration of hydronium ion is,

  $\begin{array}{l}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{=}{\text{10}}^{\text{-}\text{2}\text{.56}}\\ \text{=}\text{2}\text{.8}{\text{×10}}^{\text{-03}}\text{M}\\ \text{=}\text{0}\text{.0028}\text{M}\end{array}$

The pH value of the solution is $2.56$ which less than the pH value is $\text{7}$.  Therefore the given solution is acidic.

Interpretation Introduction

Interpretation:

The hydronium ion concentration in a solution with a $pHof9.67$ has to be determined and the solution is whether acidic or basic has to be ascertained.

Concept introduction:

• Strong acids dissociates completely into ions in solution but weak acids do not.
• pH of a solution is the negative of the base $\text{-10}$ logarithm of the hydronium ion concentration.

  $\text{pH}\text{=}\text{-log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$

• Concentration of hydronium ion $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{=}{\text{10}}^{\text{-pH}}$
• For an acidic solution $\text{pH}\text{<7}$ and for a basic solution $\text{pH>}\text{7}$.

Answer to Problem 109GQ

$\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$=$\text{2}\text{.1}{\text{×10}}^{-10}\text{M}$ and the solution is basic in nature

Explanation of Solution

Concentration of hydronium ion can be determined using the formula,

  Concentration of hydronium ion $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{=}{\text{10}}^{\text{-pH}}$

The pH of the given solution is $9.67$.  Substituting this value in the above equation,

Then concentration of hydronium ion is,

  $\begin{array}{l}\\ \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{=}{\text{10}}^{\text{-}9.67}\\ \text{=}\text{2}\text{.1}{\text{×10}}^{-10}\text{M}\end{array}$

The pH value of the solution is $9.67$ which is greater than the pH value is $\text{7}$.  Therefore the given solution is basic.

Interpretation Introduction

Interpretation:

The $pH$ of the dilute solution has to be determined.

Concept introduction:

• Strong acids dissociates completely into ions in solution but weak acids do not.
• pH of a solution is the negative of the base $\text{-10}$ logarithm of the hydronium ion concentration.

  $\text{pH}\text{=}\text{-log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$

• Concentration of hydronium ion $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{=}{\text{10}}^{\text{-pH}}$
• For an acidic solution $\text{pH}\text{<7}$ and for a basic solution $\text{pH>}\text{7}$.

Answer to Problem 109GQ

$0.99$ is the $pH$ of the dilute solution

Explanation of Solution

A $10.0mL$ sample of $2.56MHCl$ is diluted with water to $250mL$ the $pH$ of the dilute solution can be calculated as follows,

  $\begin{array}{l}AmountofHCl=Concnetration{}_{HCl}\times Volume{}_{HCl}\\ =\frac{2.56mol}{L}\times 0.010L\\ =0.0256mol\end{array}$

$\begin{array}{l}Concentrationof HCl=\frac{AmountofHCl}{VolumeofHCl}\\ =\frac{0.0256mol}{0.250}\\ =0.1024MHCl\end{array}$

  $\begin{array}{l}\text{pH}\text{=}{\text{-log}}_{\text{10}}\left[{\text{H}}^{\text{+}}\right]\\ \text{pH}\text{=}\text{-log}\text{0}\text{.1024}\\ \text{=}\text{0}\text{.99}\end{array}$