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13 February, 18:35

What mass of NaC6H5COO should be added to 1.5 L of 0.40 M C6H5COOH solution at 25 °C to produce a solution with a pH of 3.87 given that the Ka of C6H5COOH is 6.5*10-5 and the molar mass of NaC6H5COO is 144.1032 g/mol?

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  1. 13 February, 18:44
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    41 g

    Explanation:

    We have a buffer formed by a weak acid (C₆H₅COOH) and its conjugate base (C₆H₅COO⁻ coming from NaC₆H₅COO). We can find the concentration of C₆H₅COO⁻ (and therefore of NaC₆H₅COO) using the Henderson-Hasselbach equation.

    pH = pKa + log [C₆H₅COO⁻]/[C₆H₅COOH]

    pH - pKa = log [C₆H₅COO⁻] - log [C₆H₅COOH]

    log [C₆H₅COO⁻] = pH - pKa + log [C₆H₅COOH]

    log [C₆H₅COO⁻] = 3.87 - (-log 6.5 * 10⁻⁵) + log 0.40

    [C₆H₅COO⁻] = [NaC₆H₅COO] = 0.19 M

    We can find the mass of NaC₆H₅COO using the following expression.

    M = mass NaC₆H₅COO / molar mass NaC₆H₅COO * liters of solution

    mass NaC₆H₅COO = M * molar mass NaC₆H₅COO * liters of solution

    mass NaC₆H₅COO = 0.19 mol/L * 144.1032 g/mol * 1.5 L

    mass NaC₆H₅COO = 41 g
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