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Chapter 14: Problem 70

A solution is made by adding \(50.0 \mathrm{~mL}\) of \(0.200 \mathrm{M}\) acetic acid \(\left(K_{\mathrm{a}}=1.8 \times 10^{-5}\right)\) to \(50.0 \mathrm{~mL}\) of \(1.00 \times 10^{-3} \mathrm{M} \mathrm{HCl}\) a. Calculate the \(\mathrm{pH}\) of the solution. b. Calculate the acetate ion concentration.

Short Answer

Expert verified
a. The pH of the solution is 3.30. b. The acetate ion concentration is \(5.00 \times 10^{-4} \mathrm{M}\).

Step by step solution

01

Understand the reactions in the solution

When acetic acid is mixed with HCl, the stronger acid, HCl, will donate a proton to the weaker acid, acetic acid, forming acetate ion and water. The major species present in the solution now are H3O+, Cl-, and Acetate ion (CH3COO-). The reaction can be written as: CH3COOH + HCl -> CH3COO^- + H3O^+ Now, we can proceed to calculate the pH of the solution and acetate ion concentration.
02

Calculate the initial moles of Acetic Acid and HCl

To calculate the initial moles of acetic acid, we need to multiply the initial concentration and volume of acetic acid: Moles of Acetic Acid (CH3COOH) = volume × concentration = \(50.0 \mathrm{~mL} \times 0.200 \mathrm{M}\) = \(0.01 \mathrm{mol}\) Similarly, calculate the initial moles of HCl: Moles of HCl = \(50.0 \mathrm{~mL} \times 1.00 \times 10^{-3} \mathrm{M}\) = \(5.00 \times 10^{-5} \mathrm{mol}\)
03

Calculate the moles of H3O+ and Acetate ions after the reaction

From the reaction mentioned in step 1, we know that the moles of H3O+ ions formed will be equal to the moles of HCl initially present. So, moles of H3O+ = \(5.00 \times 10^{-5} \mathrm{mol}\) Similarly, the moles of Acetate ions formed will also be equal to the moles of HCl initially present. So, moles of Acetate ions = \(5.00 \times 10^{-5} \mathrm{mol}\)
04

Determine the concentration of H3O+ ions and calculate the pH

As we have found the moles of H3O+ ions, we can determine the concentration of these ions in the solution. The total volume of the solution is 100 mL after mixing. Concentration of H3O+ ions = Moles of H3O+ ions / Total volume = \(\frac{5.00 \times 10^{-5} \mathrm{mol}}{0.100 \mathrm{L}}\) = \(5.00 \times 10^{-4} \mathrm{M}\) Now, we can calculate the pH using the formula: pH = -log10[H3O+] = -log10(\(5.00 \times 10^{-4}\)) = \(3.30\) The pH of the solution is 3.30.
05

Calculate the Acetate ion concentration

To calculate the concentration of acetate ions, divide the moles of acetate ions by the total volume of the solution: Acetate ion concentration = Moles of Acetate ions / Total volume = \(\frac{5.00 \times 10^{-5} \mathrm{mol}}{0.100 \mathrm{L}}\) = \(5.00 \times 10^{-4} \mathrm{M}\) The acetate ion concentration is \(5.00 \times 10^{-4} \mathrm{M}\).

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Most popular questions from this chapter

Calculate the \(\mathrm{pH}\) of a \(0.050 \mathrm{M}\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH}\) solution \(\left(K_{\mathrm{b}}=\right.\) \(\left.1.3 \times 10^{-3}\right)\)

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