Chapter 15, Problem 1SP

(a)

Interpretation Introduction

Interpretation:

The formation of buffer solution has to be explained.

Concept Introduction:

Molarity is defined as the ratio of number of moles of solute to the volume of solution is liters.  The S.I. unit if molarity is molar and it is represented by $\text{M}$.

Mathematical formulation of molarity is shown as follows:

  $\text{Molarity}=\frac{\text{Moles of solute}}{\text{Volume of solution }\left(\text{L}\right)}$

The $\text{pH}$ of a solution is the negative logarithmic of hydrogen ion concentration in the solution.  Mathematically, it can be represented as follows:

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

Answer to Problem 1SP

The buffer solution can be made by formic acid and sodium formate containing the concentration of conjugate base which is $0.9120$ times concentration of conjugate acid.

Explanation of Solution

The buffer solution always consists of conjugate acid and conjugate base.  As per the given data the formic acid and sodium formate makes the conjugate acid base pair.

The standard value of  $K_{\text{a}}$  for formic acid used in the test is $1.8\times {10}^{-}{}^4$ .

The $\text{pH}$ of the buffered solution is $3.70$.

The ${\text{pK}}_{\text{a}}$ for formic acid is calculated by the relation shown below.

  ${\text{pK}}_{\text{a}}=-\log {\text{K}}_{\text{a}}$        (1)

The value of $K_{\text{a}}$ is $1.8\times {10}^{-}{}^4$ .

Substitute the value of  $K_{\text{a}}$  in equation (1).

  $\begin{array}{c}{\text{pK}}_{\text{a}}=-\log \left(1.8\times {10}^{-}{}^4\right)\\ =3.74\end{array}$

The Henderson-Hasselbalch equation is represented below.

  $\text{pH}={\text{pK}}_{\text{a}}+\log \frac{\left[\text{conjugate}\text{base}\right]}{\left[\text{conjugate}\text{acid}\right]}$        (2)

The value of ${\text{pK}}_{\text{a}}$ is $3.74$.

The value of $\text{pH}$ is $3.70$.

Substitute the value of ${\text{pK}}_{\text{a}}$ and $\text{pH}$ in equation (2).

  $\begin{array}{c}\text{3}\text{.70 }=3.74+\log \frac{\left[\text{conjugate}\text{base}\right]}{\left[\text{conjugate}\text{acid}\right]}\\ \log \frac{\left[\text{conjugate}\text{base}\right]}{\left[\text{conjugate}\text{acid}\right]}=3.70-3.74\\ \frac{\left[\text{conjugate}\text{base}\right]}{\left[\text{conjugate}\text{acid}\right]}={10}^{0.04}\\ =1.096\end{array}$

Thus, to make the buffer solution of formic acid and sodium formate, the concentration of conjugate base is $1.096$ times concentration of conjugate acid.

(b)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of the buffer solution after addition of hydrochloric acid has to be determined.

Concept Introduction:

Same as part (a).

Answer to Problem 1SP

The $\text{pH}$ of the resulting buffer solution is $\underset{\_}{2.28}$.

Explanation of Solution

The initial of $\text{pH}$ the buffer solution is $3.70$.

The initial hydrogen ion concentration is calculated using the formula shown below.

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

Substitute the value of $\text{pH}$ in equation ().

  $\begin{array}{c}\left[{\text{H}}^{+}\right]={10}^{-\text{3}\text{.70}}\\ =1.99\times {10}^{-4}\text{M}\end{array}$

Thus, the initial concentration of hydrogen ion $1.99\times {10}^{-4}\text{M}$.

The number of moles of hydrochloric acid added in the buffer solution is $\text{0}\text{.0050}\text{mol}$.

The concentration of hydrochloric acid is calculated using the relation shown below.

  $M=\frac{n}{V}$        (3)

Where,

• $n$ is the number of moles.
• $M$ is the molarity.
• $V$ is the volume.

The value of $n$ for $\text{HCl}$ is $\text{0}\text{.0050}\text{mol}$.

The value of $V$ for $\text{HCl}$ is $1.0\text{L}$.

Substitute the values of $n$ and $V$ for $\text{HCl}$ in the equation (3).

  $\begin{array}{c}{M}_{\text{HCl}}=\frac{0.0050\text{mol}}{1.0\text{L}}\\ =0.0050\text{M}\end{array}$

Since, hydrochloric acid is strong acid therefore it will get completely dissociated in the solution.  Therefore, the increase in the concentration of hydrogen ions is $0.0050\text{M}$.

The final concentration of hydrogen ions after addition of $\text{HCl}$ is calculated as shown below.

  $\begin{array}{c}\left[{\text{H}}^{+}\right]=1.99\times {10}^{-4}\text{M}+0.0050\text{M}\\ =\text{0}\text{.005199}\text{M}\end{array}$

The $\text{pH}$ of resulting buffer solution is calculated by using the relation shown below.

  $\text{pH}=-\log \left[{\text{H}}^{+}\right]$        (4)

The value of $\left[{\text{H}}^{+}\right]$ is $\text{0}\text{.005199}\text{M}$.

Substitute the value of $\left[{\text{H}}^{+}\right]$ in equation (4).

  $\begin{array}{c}\text{pH}=-\log \left(\text{0}\text{.005199}\right)\\ =\underset{\_}{2.28}\end{array}$

Thus, the $\text{pH}$ of the resulting buffer solution is $\underset{\_}{2.28}$.

(c)

Interpretation Introduction

Interpretation:

The mass of sodium hydroxide that is to be added to just increase the buffer capacity of initial buffer solution is to be determined.

Concept Introduction:

Same as part (a).

Answer to Problem 1SP

The mass of sodium hydroxide which is to be added is $\underset{\_}{64g}$.

Explanation of Solution

The initial buffer solution contains formic acid and sodium formate, the concentration of conjugate base is $1.096$ times concentration of conjugate acid.

The number of mole of formic acid in one liter of the solution is same as that of the concentration.  Therefore the number of moles of hydronium ion in the solution is $1.096\text{mol}$.  Thus, if the number of moles of sodium hydroxide is more than $1.096\text{mol}$, the buffer capacity gets exceeds.

The initial buffer solution cannot accommodate the addition of $1.6\text{mol}\text{NaOH}$ without undergoing a major change in $\text{pH}$.  This addition would use up all of the buffer’s conjugate acid and as the result there is increase in the buffer capacity.

The minimum mass of sodium hydroxide that is to be added to increase the buffer capacity is calculated using the relation given below.

  $n=\frac{m}{M_w}$        (5)

Where,

• $n$ is the number of moles.
• $M_w$ is the molar mass.
• $m$ is the mass.

The value of $n$ for sodium hydroxide is $1.6\text{mol}$.

The value of $M_w$ for sodium hydroxide is $40\text{g/mol}$.

Substitute the values of $n$ and $M_w$ for sodium hydroxide in the equation (5).

  $\begin{array}{c}1.6\text{mol}=\frac{m}{40\text{g/mol}}\\ m=1.6\text{mol}\times 40\text{g/mol}\\ =64\text{g}\end{array}$

Thus, the mass of sodium hydroxide which is to be added to just increase the buffer capacity is $\underset{\_}{64g}$.