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Chapter 16: Problem 56

If a solution of hydrofluoric acid $\left(\mathrm{HF} ; K_{a}=6.8 \times 10^{-4}\right)\( has a \)\mathrm{pH}$ of 2.12 , calculate the concentration of hydrofluoric acid.

Short Answer

Expert verified
The concentration of hydrofluoric acid (HF) in the solution is 0.0847 M.

Step by step solution

01

Write the dissociation of hydrofluoric acid and its equilibrium expression with Ka

Hydrofluoric acid (HF) dissociates in water as follows: HF ⇌ H+ + F- The equilibrium constant expression for this reaction is given by Ka: Ka = \(\frac{[H^+][F^-]}{[HF]}\) Since we are given the Ka value, 6.8 x 10^{-4}, we can use it in our calculations.
02

Write the definition of pH and find the H+ concentration

The pH of a solution is defined as the negative logarithm of the concentration of H+ ions in the solution: pH = -log[H+] We are given the pH value as 2.12. We can find the H+ concentration by taking the inverse logarithm of the pH value: [H+] = 10^{-2.12} Now, calculate the concentration of H+: [H+] = 7.59 x 10^{-3} M
03

Apply the ICE table (Initial, Change, Equilibrium) method

Let's set up an ICE table to find the concentration of HF at equilibrium. | HF | H+ | F- | | 1. Initial | 0 | 0 | | 2. Change | -x | +x | +x | | 3. Equilibrium | [HF]-x | [H+] +x | x | We know that [H+] = 7.59 x 10^{-3} M and that at equilibrium: [H+] + x = 7.59 x 10^{-3} x = [F-]
04

Substitute the equilibrium concentrations into the Ka expression

Now, we'll substitute the values from the ICE table into the Ka expression: \(Ka = \frac{([H^+]+x)(x)}{[HF]-x}\) Substitute the known values: \(6.8 \times 10^{-4} = \frac{(7.59 \times 10^{-3} + x)(x)}{[HF] - x}\)
05

Apply the approximation that x is very small

In weak acid/base equilibria, we can make an approximation that x is very small compared to the initial concentrations. This allows us to make a simplification in our calculations: Approximation: x << [H+] Now, we can rewrite the equation as: \(6.8 \times 10^{-4} = \frac{(7.59 \times 10^{-3})(x)}{[HF]}\)
06

Solve for the concentration of HF

Let's solve the equation for [HF]: [HF] = \(\frac{(7.59 \times 10^{-3})(x)}{6.8 \times 10^{-4}}\) Now, notice that in our ICE table, we found that x = [F-], so we can substitute [F-] for x: [HF] = \(\frac{(7.59 \times 10^{-3})([F^-])}{6.8 \times 10^{-4}}\) We also know that [F-] = [H+], so we can plug the value of [H+] in the equation: [HF] = \(\frac{(7.59 \times 10^{-3})(7.59 \times 10^{-3})}{6.8 \times 10^{-4}}\) Now, calculate the value of [HF]: [HF] = 0.0847 M The concentration of hydrofluoric acid in the solution is 0.0847 M.

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Most popular questions from this chapter

A solution is made by adding $1.000 \mathrm{~g} \mathrm{Ca}(\mathrm{OH})_{2}(s), 100.0 \mathrm{~mL}$ of \(0.10 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\), and enough water to make a final volume of \(350.0 \mathrm{~mL}\). Assuming that all of the solid dissolves, what is the \(\mathrm{pH}\) of the final solution?

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