Calculate pH for a weak acid/strong base titration.

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Calculate pH for a weak acid/strong base titration. Determine the pH during the titration of 56.9 mL of 0.301 M hydrofluoric acid (Ka = 7.2*10^4) by 0.301 M KOH at the following points: 1. Before the addition of any KOH 2. After the addition of 14.0 mL of KOH 3. At the half-equivalence point (the titration midpoint) 4. At the equivalence point 5. After the addition of 85.4 mL of KOH Check & Submit Answer Show Approach

The Correct Answer and Explanation is:

Let’s go through each step of the titration for the weak acid (hydrofluoric acid, HF) and strong base (KOH). We need to calculate the pH at different points during the titration.

Given Data:

  • Volume of HF = 56.9 mL = 0.0569 L
  • Concentration of HF = 0.301 M
  • KaK_aKa​ of HF = 7.2×10−47.2 \times 10^{-4}7.2×10−4
  • Concentration of KOH = 0.301 M
  • Volume of KOH added will vary at different points.

1. Before the addition of any KOH (Initial pH of HF)

For the initial pH of a weak acid, we need to use the expression for the dissociation of HF: HF⇌H++F−HF \rightleftharpoons H^+ + F^-HF⇌H++F−

The equilibrium expression for the dissociation is: Ka=[H+][F−][HF]K_a = \frac{[H^+][F^-]}{[HF]}Ka​=[HF][H+][F−]​

Let’s assume the concentration of [H+][H^+][H+] is xxx, and at equilibrium: Ka=x2[HF]−xK_a = \frac{x^2}{[HF] – x}Ka​=[HF]−xx2​

For weak acids, xxx will be very small compared to the initial concentration, so we can approximate [HF]−x≈[HF][HF] – x \approx [HF][HF]−x≈[HF].

Thus: Ka=x2[HF]K_a = \frac{x^2}{[HF]}Ka​=[HF]x2​

Solving for xxx, the concentration of H+H^+H+: x=Ka×[HF]x = \sqrt{K_a \times [HF]}x=Ka​×[HF]​

Substitute the values: x=(7.2×10−4)×0.301x = \sqrt{(7.2 \times 10^{-4}) \times 0.301}x=(7.2×10−4)×0.301​ x≈1.48×10−2 Mx \approx 1.48 \times 10^{-2} \, \text{M}x≈1.48×10−2M

Now, calculate the pH: pH=−log⁡([H+])=−log⁡(1.48×10−2)≈1.83\text{pH} = -\log([H^+]) = -\log(1.48 \times 10^{-2}) \approx 1.83pH=−log([H+])=−log(1.48×10−2)≈1.83

2. After the addition of 14.0 mL of KOH

To calculate the pH after adding 14.0 mL of KOH, we first determine the moles of both HF and KOH.

  • Moles of HF:

moles of HF=0.301 M×0.0569 L=0.0171 mol\text{moles of HF} = 0.301 \, \text{M} \times 0.0569 \, \text{L} = 0.0171 \, \text{mol}moles of HF=0.301M×0.0569L=0.0171mol

  • Moles of KOH:

moles of KOH=0.301 M×0.0140 L=0.00421 mol\text{moles of KOH} = 0.301 \, \text{M} \times 0.0140 \, \text{L} = 0.00421 \, \text{mol}moles of KOH=0.301M×0.0140L=0.00421mol

KOH is a strong base, and it will neutralize HF, forming F−F^-F− and H2OH_2OH2​O.

  • Remaining moles of HF:

0.0171 mol−0.00421 mol=0.0129 mol0.0171 \, \text{mol} – 0.00421 \, \text{mol} = 0.0129 \, \text{mol}0.0171mol−0.00421mol=0.0129mol

  • Moles of F−F^-F− formed:

0.00421 mol0.00421 \, \text{mol}0.00421mol

Now, calculate the new concentrations in the total volume of the solution, which is: Vtotal=56.9 mL+14.0 mL=70.9 mL=0.0709 LV_{\text{total}} = 56.9 \, \text{mL} + 14.0 \, \text{mL} = 70.9 \, \text{mL} = 0.0709 \, \text{L}Vtotal​=56.9mL+14.0mL=70.9mL=0.0709L

  • Concentration of HF:

[HF]=0.0129 mol0.0709 L=0.182 M[HF] = \frac{0.0129 \, \text{mol}}{0.0709 \, \text{L}} = 0.182 \, \text{M}[HF]=0.0709L0.0129mol​=0.182M

  • Concentration of F−F^-F−:

[F−]=0.00421 mol0.0709 L=0.0594 M[F^-] = \frac{0.00421 \, \text{mol}}{0.0709 \, \text{L}} = 0.0594 \, \text{M}[F−]=0.0709L0.00421mol​=0.0594M

At this point, we use the Henderson-Hasselbalch equation to find the pH: pH=pKa+log⁡([F−][HF])\text{pH} = \text{pKa} + \log \left( \frac{[F^-]}{[HF]} \right)pH=pKa+log([HF][F−]​)

First, calculate pKa\text{pKa}pKa: pKa=−log⁡(Ka)=−log⁡(7.2×10−4)≈3.14\text{pKa} = -\log(K_a) = -\log(7.2 \times 10^{-4}) \approx 3.14pKa=−log(Ka​)=−log(7.2×10−4)≈3.14

Then, calculate the pH: pH=3.14+log⁡(0.05940.182)≈3.14+log⁡(0.326)≈3.14−0.486≈2.65\text{pH} = 3.14 + \log \left( \frac{0.0594}{0.182} \right) \approx 3.14 + \log(0.326) \approx 3.14 – 0.486 \approx 2.65pH=3.14+log(0.1820.0594​)≈3.14+log(0.326)≈3.14−0.486≈2.65

3. At the half-equivalence point (the titration midpoint)

At the half-equivalence point, the moles of KOH added will be equal to half the moles of HF.

  • Moles of KOH at half-equivalence:

0.01712=0.00855 mol\frac{0.0171}{2} = 0.00855 \, \text{mol}20.0171​=0.00855mol

The concentration of F−F^-F− and HFHFHF will be the same at this point, so: pH=pKa=3.14\text{pH} = \text{pKa} = 3.14pH=pKa=3.14

4. At the equivalence point

At the equivalence point, all HF has reacted with KOH. The solution now contains only the conjugate base F−F^-F−, which will hydrolyze in water to form OH−OH^-OH−. The hydrolysis reaction is: F−+H2O⇌HF+OH−F^- + H_2O \rightleftharpoons HF + OH^-F−+H2​O⇌HF+OH−

We can use the Kb of F−F^-F− to calculate the pH: Kb=KwKa=1.0×10−147.2×10−4≈1.39×10−11K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{7.2 \times 10^{-4}} \approx 1.39 \times 10^{-11}Kb​=Ka​Kw​​=7.2×10−41.0×10−14​≈1.39×10−11

For the hydrolysis of F−F^-F−: Kb=[OH−]2[F−]K_b = \frac{[OH^-]^2}{[F^-]}Kb​=[F−][OH−]2​

Solving for [OH−][OH^-][OH−]: [OH−]=Kb×[F−][OH^-] = \sqrt{K_b \times [F^-]}[OH−]=Kb​×[F−]​

Substitute the values: [OH−]=(1.39×10−11)×0.301≈6.77×10−6 M[OH^-] = \sqrt{(1.39 \times 10^{-11}) \times 0.301} \approx 6.77 \times 10^{-6} \, \text{M}[OH−]=(1.39×10−11)×0.301​≈6.77×10−6M

Now, calculate the pOH: pOH=−log⁡([OH−])=−log⁡(6.77×10−6)≈5.17\text{pOH} = -\log([OH^-]) = -\log(6.77 \times 10^{-6}) \approx 5.17pOH=−log([OH−])=−log(6.77×10−6)≈5.17

Finally, the pH is: pH=14−pOH=14−5.17=8.83\text{pH} = 14 – \text{pOH} = 14 – 5.17 = 8.83pH=14−pOH=14−5.17=8.83

5. After the addition of 85.4 mL of KOH

The total volume of KOH added is 85.4 mL.

  • Moles of KOH added:

moles of KOH=0.301 M×0.0854 L=0.0257 mol\text{moles of KOH} = 0.301 \, \text{M} \times 0.0854 \, \text{L} = 0.0257 \, \text{mol}moles of KOH=0.301M×0.0854L=0.0257mol

At this point, the amount of KOH added exceeds the moles of HF, so the solution will be basic due to excess OH−OH^-OH−. The excess moles of KOH are: excess moles of KOH=0.0257 mol−0.0171 mol=0.0086 mol\text{excess moles of KOH} = 0.0257 \, \text{mol} – 0.0171 \, \text{mol} = 0.0086 \, \text{mol}excess moles of KOH=0.0257mol−0.0171mol=0.0086mol

The concentration of OH−OH^-OH− in the total volume is: total volume=56.9 mL+85.4 mL=142.3 mL=0.1423 L\text{total volume} = 56.9 \, \text{mL} + 85.4 \, \text{mL} = 142.3 \, \text{mL} = 0.1423 \, \text{L}total volume=56.9mL+85.4mL=142.3mL=0.1423L [OH−]=0.0086 mol0.1423 L=0.0605 M[OH^-] = \frac{0.0086 \, \text{mol}}{0.1423 \, \text{L}} = 0.0605 \, \text{M}[OH−]=0.1423L0.0086mol​=0.0605M

Now, calculate the pOH: pOH=−log⁡(0.0605)≈1.22\text{pOH} = -\log(0.0605) \approx 1.22pOH=−log(0.0605)≈1.22

Finally, calculate the pH: pH=14−1.22=12.78\text{pH} = 14 – 1.22 = 12.78pH=14−1.22=12.78

Summary of pH at each point:

  1. Before the addition of KOH: pH ≈ 1.83
  2. After adding 14.0 mL of KOH: pH ≈ 2.65
  3. At the half-equivalence point: pH = 3.14
  4. At the equivalence point: pH ≈ 8.83
  5. After adding 85.4 m

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