Chapter 8, Problem 66E

(a)

Interpretation Introduction

Interpretation:

The initial pH of 0.100 M H₂NNH₂ should be calculated.

Concept Introduction :

The pH of the solution can be calculated by,

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

It can also be calculated from hydroxide ion concentration as follows:

  $\text{pOH}=-\log \left[{\text{OH}}^{-}\right]$

And,

  $\text{pH}=14-\text{pOH}$

Answer to Problem 66E

  $\text{pH = 10}\text{.74}$

Explanation of Solution

  $\begin{array}{l}{\text{ H}}_2{\text{NNH}}_2{\text{ + H}}_2\text{O }\rightleftharpoons {\text{ H}}_2{\text{NNH}}_3{}^{+}{\text{ + OH}}^{-}\\ \text{initial(M) 0}\text{.100 - -}\\ \text{eqm }\left(0.100-x\right)xx\end{array}$

  $\begin{array}{l}{K}_b=\frac{\left[{\text{OH}}^{-}\right]\left[{\text{H}}_2{\text{NNH}}_3{}^{+}\right]}{\left[{\text{H}}_2{\text{NNH}}_2\right]}\\ 3.0\times {10}^{-6}=\frac{x^2}{\left(0.100-x\right)}\approx \frac{x^2}{0.100}\\ x=\left[{\text{H}}^{+}\right]=5.5\times {10}^{-4}\text{ M}\end{array}$

  $\begin{array}{l}\text{pOH = }-\log \left[{\text{OH}}^{-}\right]\\ \text{pOH = }-\log (5.5\times {10}^{-4})\\ \text{pOH = 3}\text{.26}\\ \text{pH = 14 - 3}\text{.26}\\ \text{pH = 10}\text{.74}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The pH of the resulting solution after adding 20.0 mL of 0.200 M HNO₃ to 100.0 mL of 0.100 M H₂NNH₂ should be calculated.

Concept Introduction :

Henderson-Hasselbalch is used to calculate the pH of a buffer solution and the equation is represented as follow:

  ${\text{pH = pK}}_a+\log \frac{\left[\text{conjugate base}\right]}{\left[\text{acid}\right]}$

Answer to Problem 66E

  $\text{pH = 8}\text{.66}$

Explanation of Solution

The added H+ reacts completely with hydrazine.

  ${\text{H}}_2{\text{NNH}}_2{\text{ + H}}^{+}\to {\text{H}}_2{\text{NNH}}_3^{+}{\text{ + OH}}^{-}$

Number of moles of hydrazine present = $\frac{0.100\text{ mol}}{1000\text{ mL}}\times 100.0\text{ mL = 0}\text{.01 mol}$

Number of moles of H+ added = $\frac{0.200\text{ mol}}{1000\text{ mL}}\times 20.0\text{ mL = 0}\text{.004 mol}$

Number of moles of hydrazine left = $0.01-0.004=0.006\text{ mol}$

Number of moles of conjugate acid formed = $0.004\text{ mol}$

  $K_a=\frac{K_w}{K_b}=\frac{1.0\times {10}^{-14}}{3.0\times {10}^{-6}}=3.3\times {10}^{-9}$

If the final volume is V,

  $\begin{array}{l}{\text{pH = pK}}_a+\log \frac{\left[{\text{H}}_2{\text{NNH}}_2\right]}{\left[{\text{H}}_2{\text{NNH}}_3{}^{+}\right]}\\ \text{pH = }-\log \left(3.3\times {10}^{-9}\right)+\log \left(\frac{0.006/V}{0.004/V}\right)\\ \text{pH = 8}\text{.48 + 0}\text{.18}\\ \text{pH = 8}\text{.66}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The pH of the resulting solution after adding 25.0 mL of 0.200 M HNO₃ to 100.0 mL of 0.100 M H₂NNH₂ should be calculated.

Concept Introduction :

Henderson-Hasselbalch is used to calculate the pH of a buffer solution and the equation is represented as follow:

  ${\text{pH = pK}}_a+\log \frac{\left[\text{conjugate base}\right]}{\left[\text{acid}\right]}$

Answer to Problem 66E

  $\text{pH = 8}\text{.48}$

Explanation of Solution

  ${\text{H}}_2{\text{NNH}}_2{\text{ + H}}^{+}\to {\text{H}}_2{\text{NNH}}_3^{+}{\text{ + OH}}^{-}$

Number of moles of hydrazine present = $\frac{0.100\text{ mol}}{1000\text{ mL}}\times 100.0\text{ mL = 0}\text{.01 mol}$

Number of moles of H+ added = $\frac{0.200\text{ mol}}{1000\text{ mL}}\times 25.0\text{ mL = 0}\text{.005 mol}$

Number of moles of hydrazine left = $0.01-0.005=0.005\text{ mol}$

Number of moles of conjugate acid formed = $0.005\text{ mol}$

  $K_a=\frac{K_w}{K_b}=\frac{1.0\times {10}^{-14}}{3.0\times {10}^{-6}}=3.3\times {10}^{-9}$

If the final volume is V,

  $\begin{array}{l}{\text{pH = pK}}_a+\log \frac{\left[{\text{H}}_2{\text{NNH}}_2\right]}{\left[{\text{H}}_2{\text{NNH}}_3{}^{+}\right]}\\ \text{pH = }-\log \left(3.3\times {10}^{-9}\right)+\log \left(\frac{0.005/V}{0.005/V}\right)\\ \text{pH = 8}\text{.48}\end{array}$

(d)

Interpretation Introduction

Interpretation:

The pH of the resulting solution after adding 40.0 mL of 0.200 M HNO₃ to 100.0 mL of 0.100 M H₂NNH₂ should be calculated.

Concept Introduction :

Henderson-Hasselbalch is used to calculate the pH of a buffer solution and the equation is represented as follow:

  ${\text{pH = pK}}_a+\log \frac{\left[\text{conjugate base}\right]}{\left[\text{acid}\right]}$

Answer to Problem 66E

  $\text{pH = 7}\text{.88}$

Explanation of Solution

  ${\text{H}}_2{\text{NNH}}_2{\text{ + H}}^{+}\to {\text{H}}_2{\text{NNH}}_3^{+}{\text{ + OH}}^{-}$

Number of moles of hydrazine present = $\frac{0.100\text{ mol}}{1000\text{ mL}}\times 100.0\text{ mL = 0}\text{.01 mol}$

Number of moles of H+ added = $\frac{0.200\text{ mol}}{1000\text{ mL}}\times 40.0\text{ mL = 0}\text{.008 mol}$

Number of moles of hydrazine left = $0.01-0.008=0.002\text{ mol}$

Number of moles of conjugate acid formed = $0.008\text{ mol}$

  $K_a=\frac{K_w}{K_b}=\frac{1.0\times {10}^{-14}}{3.0\times {10}^{-6}}=3.3\times {10}^{-9}$

If the final volume is V,

  $\begin{array}{l}{\text{pH = pK}}_a+\log \frac{\left[{\text{H}}_2{\text{NNH}}_2\right]}{\left[{\text{H}}_2{\text{NNH}}_3{}^{+}\right]}\\ \text{pH = }-\log \left(3.3\times {10}^{-9}\right)+\log \left(\frac{0.002/V}{0.008/V}\right)\\ \text{pH = 8}\text{.48 + }\left(-0.60\right)\\ \text{pH = 7}\text{.88}\end{array}$

(e)

Interpretation Introduction

Interpretation:

The pH of the resulting solution after adding 50.0 mL of 0.200 M HNO₃ to 100.0 mL of 0.100 M H₂NNH₂ should be calculated.

Concept Introduction :

Henderson-Hasselbalch is used to calculate the pH of a buffer solution and the equation is represented as follow:

  ${\text{pH = pK}}_a+\log \frac{\left[\text{conjugate base}\right]}{\left[\text{acid}\right]}$

Answer to Problem 66E

  $\text{pH = 4}\text{.82}$

Explanation of Solution

  ${\text{H}}_2{\text{NNH}}_2{\text{ + H}}^{+}\to {\text{H}}_2{\text{NNH}}_3^{+}{\text{ + OH}}^{-}$

Number of moles of hydrazine present = $\frac{0.100\text{ mol}}{1000\text{ mL}}\times 100.0\text{ mL = 0}\text{.01 mol}$

Number of moles of H+ added = $\frac{0.200\text{ mol}}{1000\text{ mL}}\times 50.0\text{ mL = 0}\text{.01 mol}$

Number of moles of hydrazine left = $0.01-0.01=0.0\text{ mol}$

Number of moles of conjugate acid formed = $0.01\text{ mol}$

This is the equivalence point of the titration. pH at the equivalence point of a weak base-strong acid titration, solely comes from the conjugate acid formed.

  $\begin{array}{l}{\text{ H}}_2{\text{NNH}}_3^{+}\rightleftharpoons {\text{ H}}^{+}{\text{ + H}}_2{\text{NNH}}_2\\ \text{initial 0}\text{.01 mol/150}\text{.0 mL - -}\\ \text{eqm }\left(0.0667-x\right)xx\end{array}$

  $\begin{array}{l}{K}_a=\frac{\left[{\text{H}}^{+}\right]\left[{\text{H}}_2{\text{NNH}}_2\right]}{\left[{\text{H}}_2{\text{NNH}}_3^{+}\right]}\\ 3.3\times {10}^{-9}=\frac{x^2}{0.0667-x}\approx \frac{x^2}{0.0667}\\ x=\left[{\text{H}}^{+}\right]=1.5\times {10}^{-5}\text{ M}\end{array}$

  $\begin{array}{l}\text{pH = }-\log \left[{\text{H}}^{+}\right]\\ \text{pH = }-\log \left(1.5\times {10}^{-5}\right)\\ \text{pH = 4}\text{.82}\end{array}$

(f)

Interpretation Introduction

Interpretation:

The pH of the resulting solution after adding 100.0 mL of 0.200 M HNO₃ to 100.0 mL of 0.100 M H₂NNH₂ should be calculated.

Concept Introduction :

Henderson-Hasselbalch is used to calculate the pH of a buffer solution and the equation is represented as follow:

  ${\text{pH = pK}}_a+\log \frac{\left[\text{conjugate base}\right]}{\left[\text{acid}\right]}$

Answer to Problem 66E

  $\text{pH = 1}\text{.3}$

Explanation of Solution

Number of moles of H+ added = $\frac{0.200\text{ mol}}{1000\text{ mL}}\times 100.0\text{ mL = 0}\text{.02 mol}$

Number of moles of conjugate base formed = $0.01\text{ mol}$

Number of excess H+ moles from HNO₃ = $0.02-0.01=0.01\text{ mol}$

When a weak acid and a strong acid both are present in a solution, the amount of H+ added from weak acid can be neglected as its dissociation constant is very small. The pH past the equivalence point is determined by the amount of excess acid added.

  ${\left[{\text{H}}^{+}\right]}_{excess}=\frac{0.01\text{ mol}}{\left(100.0+100.0\right)\text{ mL}}=0.05\text{ M}$

  $\begin{array}{l}\text{pH = }-\log \left[{\text{H}}^{+}\right]\\ \text{pH = }-\log \left(0.05\right)\\ \text{pH = 1}\text{.3}\end{array}$