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Chapter 13: Problem 81

One mole of a weak acid HA was dissolved in 2.0 L of solution. After the system had come to equilibrium, the concentration of HA was found to be 0.45 \(M .\) Calculate \(K_{\mathrm{a}}\) for HA.

Short Answer

Expert verified
The equilibrium constant, Ka, for the weak acid HA can be calculated using the expression: Ka = \([H^+][A^-] / [HA]\) Given the equilibrium concentrations [HA]eq = 0.45 M, [H+]eq = 0.05 M, and [A-]eq = 0.05 M, plug them into the expression: Ka = \((0.05)(0.05) / 0.45\) Ka ≈ 0.00556

Step by step solution

01

Write down the equilibrium expression for the dissociation of HA

The equilibrium expression for the dissociation of a weak acid HA is: HA(aq) ⇌ H+(aq) + A-(aq)
02

Calculate the initial concentration of HA

We are given that one mole of HA is dissolved in 2.0 L of solution. So, the initial concentration of HA is: Initial concentration of HA = Moles of HA / Volume of solution Initial concentration of HA = 1 mol / 2.0 L Initial concentration of HA = 0.5 M
03

Determine the change in concentration at equilibrium

As the equilibrium concentration of HA is found to be 0.45 M, the change in the concentration of HA is: ∆[HA] = [HA]initial - [HA]eq ∆[HA] = 0.5 M - 0.45 M ∆[HA] = 0.05 M Since HA dissociates into H+ and A-, the change in their concentrations at equilibrium is equal to the change in HA concentration: ∆[H+] = ∆[A-] = 0.05 M
04

Write down the equilibrium concentrations for all species

The equilibrium concentrations of all species are: [HA]eq = 0.45 M (Given) [H+]eq = ∆[H+] = 0.05 M [A-]eq = ∆[A-] = 0.05 M
05

Calculate Ka

Now, we can write the equilibrium constant expression, Ka, for the weak acid HA as: Ka = \([H^+][A^-] / [HA]\) Substitute the equilibrium concentrations into the Ka expression: Ka = \((0.05)(0.05) / 0.45\) Ka = \(\frac{0.0025}{0.45}\) After calculating Ka, we get: Ka ≈ 0.00556

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Most popular questions from this chapter

Calculate the \(\mathrm{pH}\) of each of the following solutions of a strong acid in water. a. \(0.10 \mathrm{M} \mathrm{HCl}\) b. 5.0 M HCl c. \(1.0 \times 10^{-11} \mathrm{M} \mathrm{HCl}\)

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