Chapter 6, Problem 6G.12E

(a)

Interpretation Introduction

Interpretation:

The $pH$ of the resulting solution after addition of a strong base and the change in $pH$ has to be calculated.

Concept Introduction:

Henderson – Hasselbalch equation:

The $pH$ of a buffer can be easily calculated by using Henderson – Hasselbalch equation as given below.

  $pH=p{K}_a+\log \frac{{\left[base\right]}_{initial}}{{\left[acid\right]}_{initial}}$

Explanation of Solution

Given that, the $100mL$ buffer contains $0.140MN{a}_{2}HP{O}_4\left(aq\right)$ and $0.120MK{H}_{2}P{O}_4\left(aq\right).$

By using Henderson – Hasselbalch equation, the initial $pH$ of the above buffer solution can be calculated.

  $\begin{array}{c}pH=p{K}_a+\log \frac{{\left[base\right]}_{initial}}{{\left[acid\right]}_{initial}}\\ \\ pH=7.21+\log \frac{0.140M}{0.120M}\\ \\ =7.28\end{array}$

The addition of a base will react with the weak acid and thus decreases the concentration of weak acid and increases the concentration of conjugate base.

The reaction of added base with some of $K{H}_{2}P{O}_4\left(aq\right)$ is given below.

  $H_{2}P{O}_4{}^{-}\left(aq\right)+O{H}^{-}\left(aq\right)\to HP{O}_4{}^{2-}\left(aq\right)+H_{2}O\left(l\right)$

The initial number of moles of acid can be calculated as shown below.

  $\begin{array}{c}n{\left(H_{2}P{O}_4{}^{-}\right)}_{initial}=0.100L\times 0.120mol.L^{-1}\\ \\ =0.012mol\end{array}$

The number of moles of added base can be calculated as shown below.

  $\begin{array}{c}n{\left(NaOH\right)}_{initial}=0.075L\times 0.010mol.L^{-1}\\ \\ =0.0075mol\end{array}$

The molar ratio of acid and added base in the above equation is $1:1.$  The number of reacted moles of acid after addition of $0.0075molofNaOH$ can be calculated as given below.

  $\begin{array}{c}n{\left(H_{2}P{O}_4{}^{-}\right)}_{reacts}=0.0075molO{H}^{-}\times \frac{1mol{H}_{2}P{O}_4{}^{-}}{1molO{H}^{-}}\\ \\ =0.0075mol{H}_{2}P{O}_4{}^{-}\end{array}$

Now, the remaining amount of acid can be estimated.

  $\begin{array}{c}{n}_{final}=n_{initial}-n_{reacts}\\ \\ =0.012-0.0075\\ \\ =4.5\times {10}^{-3}mol\\ \\ \left[H_{2}P{O}_4{}^{-}\right]=\frac{Moles}{Volume}=\frac{4.5\times {10}^{-3}mol}{\left(0.100+0.075\right)L}=0.026M\end{array}$

The new concentration of conjugate base can be calculated as shown below.

  $\begin{array}{c}n{\left(HP{O}_4{}^{2-}\right)}_{initial}=0.100L\times 0.140mol.L^{-1}\\ \\ =0.014mol\\ \\ n{\left(HP{O}_4{}^{2-}\right)}_{final}=0.014mol+0.0075mol\\ \\ =0.0215mol\\ \\ \left[HP{O}_4{}^{2-}\right]=\frac{Moles}{Volume}=\frac{0.0215mol}{\left(0.100+0.075\right)L}=0.123M\end{array}$

Now, the $pH$ of the buffer solution can be calculated as given below.

  $\begin{array}{c}pH=p{K}_a+\log \frac{{\left[base\right]}_{initial}}{{\left[acid\right]}_{initial}}\\ \\ pH=7.21+\log \frac{0.123M}{0.026M}\\ \\ \text{=}7.88\\ \\ ChangeinpH=Final-initial=7.88-7.28=0.6\end{array}$

Therefore, the $pH$ of the resulting solution and the change in $pH$ is $7.88and0.6$ respectively.

(b)

Interpretation Introduction

Interpretation:

The $pH$ of the resulting solution after addition of a strong acid and the change in $pH$ has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Given that, the $100mL$ buffer contains $0.140MN{a}_{2}HP{O}_4\left(aq\right)$ and $0.120MK{H}_{2}P{O}_4\left(aq\right).$

By using Henderson – Hasselbalch equation, the initial $pH$ of the above buffer solution can be calculated.

  $\begin{array}{c}pH=p{K}_a+\log \frac{{\left[base\right]}_{initial}}{{\left[acid\right]}_{initial}}\\ \\ pH=7.21+\log \frac{0.140M}{0.120M}\\ \\ =7.28\end{array}$

The addition of a strong acid will react with the conjugate base and thus decreases the concentration of conjugate base and increases the concentration of weak acid.

The reaction of added acid with some of $HP{O}_4{}^{2-}$ is given below.

  $HP{O}_4{}^{2-}\left(aq\right)+H^{+}\left(aq\right)\to H_{2}P{O}_4{}^{-}\left(aq\right)$

The initial number of moles of acid can be calculated as shown below.

  $\begin{array}{c}n{\left(H_{2}P{O}_4{}^{-}\right)}_{initial}=0.100L\times 0.120mol.L^{-1}\\ \\ =0.012mol\end{array}$

The number of moles of added acid can be calculated as shown below.

  $\begin{array}{c}n{\left(HN{O}_3\right)}_{initial}=0.01L\times 0.50mol.L^{-1}\\ \\ =0.005mol\end{array}$

The new concentration of weak acid can be calculated as shown below.

  $\begin{array}{c}n{\left(H_{2}P{O}_4{}^{-}\right)}_{final}=0.012mol+0.005mol\\ \\ =0.017mol\\ \\ \left[H_{2}P{O}_4{}^{-}\right]=\frac{Moles}{Volume}=\frac{0.017mol}{\left(0.100+0.01\right)L}=0.1545M\end{array}$

The molar ratio of conjugate base and added acid is $1:1.$  The number of reacted moles of conjugate base after addition of $0.005molofHN{O}_3$ can be calculated as given below.

  $\begin{array}{c}n{\left(HP{O}_4{}^{2-}\right)}_{reacts}=0.005mol{H}^{+}\times \frac{1molHP{O}_4{}^{2-}}{1mol{H}^{+}}\\ \\ =0.005molHP{O}_4{}^{2-}\end{array}$

Now, the remaining amount of conjugate base can be estimated.

  $\begin{array}{c}n{\left(HP{O}_4{}^{2-}\right)}_{initial}=0.100M\times 0.140L\\ \\ =0.014mol\\ \\ n_{final}=n_{initial}-n_{reacts}\\ \\ =0.014-0.005\\ \\ =0.009mol\\ \\ \left[HP{O}_4{}^{2-}\right]=\frac{Moles}{Volume}=\frac{0.009mol}{\left(0.100+0.01\right)L}=0.082M\end{array}$

Now, the $pH$ of the buffer solution can be calculated as given below.

  $\begin{array}{c}pH=p{K}_a+\log \frac{{\left[base\right]}_{initial}}{{\left[acid\right]}_{initial}}\\ \\ pH=7.21+\log \frac{0.082M}{0.1545M}\\ \\ =6.93\\ \\ ChangeinpH=Final-initial=6.93-7.21=-0.28\left(adecreaseof0.28\right)\end{array}$

Therefore, the $pH$ of the resulting solution and the change in $pH$ is $6.93and-0.28$ respectively.