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Chapter 14: Problem 67

You have \(75.0 \mathrm{mL}\) of \(0.10 \mathrm{M}\) HA. After adding \(30.0 \mathrm{mL}\) of \(0.10 M \mathrm{NaOH},\) the \(\mathrm{pH}\) is \(5.50 .\) What is the \(K_{\mathrm{a}}\) value of \(\mathrm{HA} ?\)

Short Answer

Expert verified
The Ka value of the weak acid HA is approximately \(2.08 \times 10^{-5}\).

Step by step solution

01

Calculate initial moles of HA and NaOH

Calculate the initial moles of HA and NaOH before the reaction. For HA: moles = Molarity × Volume moles of HA = 0.10 M × 75.0 mL = 0.10 M × 0.075 L = 0.0075 moles For NaOH: moles = Molarity × Volume moles of NaOH = 0.10 M × 30.0 mL = 0.10 M × 0.030 L = 0.0030 moles
02

Perform the reaction between HA and NaOH and determine the remaining moles of HA

Since NaOH is a strong base, it will react completely with the weak acid HA. Determine the remaining moles of HA after the reaction. moles of HA remaining = initial moles of HA - moles of NaOH = 0.0075 - 0.0030 = 0.0045 moles
03

Calculate the final volume of the solution

Determine the final volume of the solution by adding the initial volumes of both HA and NaOH solutions. Final volume = initial volume of HA + initial volume of NaOH = 75.0 mL + 30.0 mL = 105.0 mL = 0.105 L
04

Calculate the equilibrium concentration of HA

Determine the equilibrium concentration of HA by dividing its remaining moles by the final volume of the solution. \[C_{HA, eq} = \frac{moles\:of\:HA\: remaining}{final\:volume} = \frac{0.0045\: moles}{0.105\: L} = 0.0429 M\]
05

Use the pH to find the equilibrium concentration of H+ ions

We are given that the pH at equilibrium is 5.50. Use the formula: pH = -log10[H+] Solve for [H+]: \[[H^{+}] = 10^{-pH} = 10^{-5.50} = 3.16 \times 10^{-6} M\]
06

Determine the equilibrium concentration of A- ions

Since the moles of NaOH completely reacted with HA, the moles of A- ions produced at equilibrium are the same as the initial moles of NaOH. Calculate the equilibrium concentration of A- ions by dividing the moles of A- ions by the final volume of the solution. \[C_{A^-, eq} = \frac{moles\:of\:A^-}{final\:volume} = \frac{0.0030\: moles}{0.105\: L} = 0.0286 M\]
07

Calculate the Ka value using equilibrium concentrations

Use the ion concentrations at equilibrium to find the Ka value of HA using the formula: \[K_{a} = \frac{[H^{+}][A^{-}]}{[HA]_{eq}}\] Substitute the equilibrium concentrations found in previous steps: \[K_{a} = \frac{(3.16 \times 10^{-6})(0.0286)}{0.0429} = 2.08 \times 10^{-5}\] The Ka value of the weak acid HA is approximately \(2.08 \times 10^{-5}\).

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Most popular questions from this chapter

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