Calculate the pH of the solution after the addition of 10.0 ml of NaOH solution.

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20.0 ml of 0.200M hypobromous acid, HBrO, is titrated with 0.250M sodium hydroxide, NaOH. Calculate the pH of the solution after the addition of 10.0 ml of NaOH solution.

Then Determine

25.0 ml of 0.600M hypobromous acid, HBrO, is titrated with 0.400M sodium hydroxide, NaOH. Calculate the pH of the solution at the equivalence point of the titration.

Hypobromous acid Ka=2.3E-9

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The Correct Answer and Explanation is:

Answer Summary

  1. After adding 10.0 mL of NaOH to 20.0 mL of 0.200 M HBrO,
    pH ≈ 8.66
  2. At the equivalence point of titrating 25.0 mL of 0.600 M HBrO with 0.400 M NaOH,
    pH ≈ 10.34

Detailed Explanation

Hypobromous acid (HBrO) is a weak acid, and NaOH is a strong base. In acid-base titrations involving a weak acid and a strong base, the pH depends on the stage of the titration.

1. After Adding 10.0 mL of NaOH to 20.0 mL of 0.200 M HBrO

Step 1: Moles of Reactants

  • Moles of HBrO = 0.0200 L × 0.200 mol/L = 0.00400 mol
  • Moles of NaOH = 0.0100 L × 0.250 mol/L = 0.00250 mol

Step 2: Reaction
HBrO+OH−→BrO−+H2O\text{HBrO} + \text{OH}^- \rightarrow \text{BrO}^- + H_2O

  • NaOH reacts with HBrO to form BrO⁻.
  • Moles of HBrO remaining = 0.00400 – 0.00250 = 0.00150 mol
  • Moles of BrO⁻ formed = 0.00250 mol
  • Total volume = 20.0 + 10.0 = 30.0 mL = 0.0300 L

Step 3: Use Henderson-Hasselbalch Equation pKa=−log⁡(2.3×10−9)≈8.64\text{pKa} = -\log(2.3 \times 10^{-9}) ≈ 8.64 pH=pKa+log⁡([BrO−][HBrO])=8.64+log⁡(0.002500.00150)=8.64+log⁡(1.6667)≈8.64+0.22=8.66\text{pH} = pKa + \log\left( \frac{[\text{BrO}^-]}{[\text{HBrO}]} \right) = 8.64 + \log\left( \frac{0.00250}{0.00150} \right) = 8.64 + \log(1.6667) ≈ 8.64 + 0.22 = \boxed{8.66}

2. At the Equivalence Point of Titrating 25.0 mL of 0.600 M HBrO with 0.400 M NaOH

Step 1: Moles of HBrO = 0.025 L × 0.600 mol/L = 0.0150 mol
Since NaOH is in stoichiometric proportion, 0.0150 mol is required.

Volume of NaOH needed: V=0.0150 mol0.400 mol/L=0.0375 L=37.5 mLV = \frac{0.0150\ \text{mol}}{0.400\ \text{mol/L}} = 0.0375\ \text{L} = 37.5\ \text{mL}

Total volume = 25.0 + 37.5 = 62.5 mL = 0.0625 L

At equivalence, all HBrO is converted to BrO⁻ (a weak base).

[BrO⁻] = 0.0150 mol / 0.0625 L = 0.240 M

Now calculate pH using base hydrolysis: Kb=KwKa=1.0×10−142.3×10−9=4.35×10−6\text{Kb} = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{2.3 \times 10^{-9}} = 4.35 \times 10^{-6} [OH−]=Kb×[BrO−]=4.35×10−6×0.240≈1.044×10−6≈1.02×10−3[OH^-] = \sqrt{Kb \times [BrO^-]} = \sqrt{4.35 \times 10^{-6} \times 0.240} ≈ \sqrt{1.044 \times 10^{-6}} ≈ 1.02 \times 10^{-3} pOH=−log⁡(1.02×10−3)≈2.99⇒pH=14−2.99=10.34\text{pOH} = -\log(1.02 \times 10^{-3}) ≈ 2.99 \quad \Rightarrow \quad \text{pH} = 14 – 2.99 = \boxed{10.34}

✅ Final Answers:

  • pH after 10.0 mL NaOH added: 8.66
  • pH at equivalence point: 10.34

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