Acid-Base Titration Notes

Acid-Base Equilibrium Practice #5

Titrations - Problem 1

• Scenario: Titration of 35.0 mL of a 0.225 M solution of potassium hydroxide (KOH) with 0.265 M butyric acid (HBut, $HC<em>4H</em>7O<em>2$). The acid dissociation constant for butyric acid is given as $K</em>a = 1.50 \times 10^{-5}$.

Part a: Balanced Chemical Equation

• The balanced chemical equation for the reaction between potassium hydroxide (KOH) and butyric acid ($HC<em>4H</em>7O<em>2$) is: $KOH + HC</em>4H<em>7O</em>2 \rightleftharpoons KC<em>4H</em>7O<em>2 + H</em>2O$
  Here, $HC<em>4H</em>7O<em>2$ represents butyric acid (HBut) and $KC</em>4H<em>7O</em>2$ represents potassium butyrate (KBut).

Part b: Volume of Butyric Acid at Equivalence Point

• To find the volume of butyric acid needed to reach the equivalence point, we use the relationship: $M<em>1V</em>1 = M<em>2V</em>2$, where $M<em>1$ and $V</em>1$ are the molarity and volume of KOH, and $M<em>2$ and $V</em>2$ are the molarity and volume of butyric acid.

• Given:

  • $M_1$ (KOH) = 0.225 M

  • $V_1$ (KOH) = 35.0 mL = 0.035 L

  • $M_2$ (Butyric acid) = 0.265 M

• Solving for $V<em>2$: $V</em>2 = \frac{M<em>1V</em>1}{M_2} = \frac{(0.225 \text{ M})(0.035 \text{ L})}{(0.265 \text{ M})} = 0.0297 \text{ L}$

• Therefore, the volume of butyric acid solution needed to reach the equivalence point is 29.7 mL.

Part c: pH After Adding 25.0 mL of Butyric Acid

• First, calculate the moles of KOH and $HC<em>4H</em>7O_2$:

  • Moles of KOH = $0.225 \text{ M} \times 0.035 \text{ L} = 0.007875 \text{ mol}$

  • Moles of $HC<em>4H</em>7O_2$ = $0.265 \text{ M} \times 0.025 \text{ L} = 0.006625 \text{ mol}$

• Subtract the moles of $HC<em>4H</em>7O_2$ from KOH to find the remaining moles of KOH:

  • Remaining moles of KOH = $0.007875 \text{ mol} - 0.006625 \text{ mol} = 0.00125 \text{ mol}$

• Calculate the concentration of remaining KOH:

  • Total volume = 35.0 mL + 25.0 mL = 60.0 mL = 0.060 L

  • $[KOH] = \frac{0.00125 \text{ mol}}{0.060 \text{ L}} = 0.0208 \text{ M}$

• Calculate the concentration of $C<em>4H</em>7O_2$:

  • $[C<em>4H</em>7O_2] = \frac{0.006625 \text{ mol}}{0.060 \text{ L}} = 0.1104 \text{ M}$

• Calculate the pKa:

  • $pK<em>a = -\log(K</em>a) = -\log(1.5 \times 10^{-5}) = 4.82$

• Calculate the pKb:

  • $pK<em>b = 14.00 - pK</em>a = 14.00 - 4.82 = 9.18$

• Calculate the pOH using the Henderson-Hasselbalch equation:

  • $pOH = pK<em>b + \log\frac{[KOH]}{[C</em>4H<em>7O</em>2^-]} = 9.18 + \log\frac{0.006625}{0.00125} = 8.457$

• Calculate the pH:

  • $pH = 14.00 - pOH = 14.00 - 8.457 = 5.543 \approx 5.5$

Titrations - Problem 2

• Scenario: Titration of 50.0 mL of a 0.235 M solution of methylamine ($CH<em>3NH</em>2$), a weak base with $K<em>b = 4.4 \times 10^{-4}$, with a 0.300 M solution of nitric acid ($HNO</em>3$).

Part a: Balanced Chemical Equation

• The balanced chemical equation for the reaction between nitric acid and methylamine is:
  $HNO<em>3 + CH</em>3NH<em>2 \rightleftharpoons CH</em>3NH<em>3^+ + NO</em>3^-$

Part b: Volume of Nitric Acid at Equivalence Point

• To find the volume of nitric acid needed to reach the equivalence point, we use the relationship: $M<em>1V</em>1 = M<em>2V</em>2$, where $M<em>1$ and $V</em>1$ are the molarity and volume of methylamine, and $M<em>2$ and $V</em>2$ are the molarity and volume of nitric acid.

• Given:

  • $M_1$ (Methylamine) = 0.235 M

  • $V_1$ (Methylamine) = 50.0 mL = 0.050 L

  • $M_2$ (Nitric acid) = 0.300 M

• Solving for $V<em>2$: $V</em>2 = \frac{M<em>1V</em>1}{M_2} = \frac{(0.235 \text{ M})(0.050 \text{ L})}{(0.300 \text{ M})} = 0.0392 \text{ L}$

• Therefore, the volume of nitric acid solution needed to reach the equivalence point is 39.2 mL.

Part c: pH After Adding 20.00 mL of Nitric Acid

• First, calculate the moles of $HNO<em>3$ and $CH</em>3NH_2$:

  • Moles of $HNO_3$ = $0.300 \text{ M} \times 0.020 \text{ L} = 0.006 \text{ mol}$

  • Moles of $CH<em>3NH</em>2$ = $0.235 \text{ M} \times 0.050 \text{ L} = 0.01175 \text{ mol}$

• Subtract the moles of $HNO<em>3$ from $CH</em>3NH<em>2$ to find the remaining moles of $CH</em>3NH_2$:

  • Remaining moles of $CH<em>3NH</em>2$ = $0.01175 \text{ mol} - 0.006 \text{ mol} = 0.00575 \text{ mol}$

• Calculate the moles of $CH<em>3NH</em>3^+$ formed:

  • Moles of $CH<em>3NH</em>3^+$ = 0.006 mol (since it's formed from the reaction with $HNO_3$)

• Calculate the concentrations of $CH<em>3NH</em>2$ and $CH<em>3NH</em>3^+$:

  • Total volume = 50.0 mL + 20.0 mL = 70.0 mL = 0.070 L

  • $[CH<em>3NH</em>2] = \frac{0.00575 \text{ mol}}{0.070 \text{ L}} = 0.0821 \text{ M}$

  • $[CH<em>3NH</em>3^+] = \frac{0.006 \text{ mol}}{0.070 \text{ L}} = 0.0857 \text{ M}$

• Calculate the pKb:

  • $pK<em>b = -\log(K</em>b) = -\log(4.4 \times 10^{-4}) = 3.357$

• Calculate the pOH using the Henderson-Hasselbalch equation:

  • $pOH = pK<em>b + \log\frac{[CH</em>3NH<em>3^+]}{[CH</em>3NH_2]} = 3.357 + \log\frac{0.0857 \text{ M}}{0.0821 \text{ M}} = 3.3755$

• Calculate the pH:

  • $pH = 14.00 - pOH = 14.00 - 3.3755 = 10.6245 \approx 10.6$

Part d: pH After Adding 50.00 mL of Nitric Acid

• Moles of $HNO_3$:

  • $0.300 \text{ M} \times 0.050 \text{ L} = 0.015 \text{ mol}$

• Moles of $CH<em>3NH</em>2$:

  • $0.235 \text{ M} \times 0.050 \text{ L} = 0.01175 \text{ mol}$

• Since the moles of $HNO<em>3$ exceed the moles of $CH</em>3NH<em>2$, all of the $CH</em>3NH<em>2$ will be consumed and we'll have an excess of $HNO</em>3$.

• Excess moles of $HNO_3$:

  • $0.015 \text{ mol} - 0.01175 \text{ mol} = 0.00325 \text{ mol}$

• Total volume:

  • 50 mL + 50 mL = 100 mL = 0.100 L

Part e: pH at the Equivalence Point

• Equation: $CH<em>3NH</em>3 \rightleftharpoons CH<em>3NH</em>2 + H_2O$