Multiple Choice
A 60.0 mL solution of 0.115 M sodium hypochlorite (NaOCl) is titrated with 0.222 M NaOH. Calculate the pH of the solution after 6.05 mL of NaOH is added. The pKa of hypochlorous acid (HOCl) is 7.53.
18. Aqueous Equilibrium
Acid-Base Indicators
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During the titration of 26.00 mL of 0.400 M formic acid with 0.380 M NaOH, what is the pH of the solution at the half-equivalence point?
A
4.00
B
7.00
C
2.50
D
3.74
Verified step by step guidance1
Identify that the problem involves a titration of a weak acid (formic acid) with a strong base (NaOH). At the half-equivalence point, the concentration of the acid equals the concentration of its conjugate base.
Recall that at the half-equivalence point, the pH of the solution is equal to the pKa of the weak acid. This is because the concentrations of the acid and its conjugate base are equal, simplifying the Henderson-Hasselbalch equation to pH = pKa.
Determine the pKa of formic acid. The pKa is related to the acid dissociation constant (Ka) by the equation: \( \text{pKa} = -\log_{10}(\text{Ka}) \). Look up the Ka value for formic acid, which is approximately \( 1.8 \times 10^{-4} \).
Calculate the pKa using the formula: \( \text{pKa} = -\log_{10}(1.8 \times 10^{-4}) \). This will give you the pH at the half-equivalence point.
Conclude that the pH at the half-equivalence point is equal to the pKa of formic acid, which is approximately 3.74.
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