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15 April, 07:13

Calculate the pH after 0.16 mole of NaOH is added to 1.06 L of a solution that is 0.48 M HF and 1.12 M NaF, and calculate the pH after 0.32 mole of HCl is added to 1.06 L of the same solution of HF and NaF.

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  1. 15 April, 09:10
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    pH = 3.73 for first part

    pH = 3.16 for second part

    Explanation:

    Use the Henderson-Hasselbalch equation to calculate the pH of this buffer solution:

    pH = pKa + log ((A⁻) / (HA))

    where pka for HF is 3.14

    Now we need to calculate the reaction of NaOH with HF in the first part and the reaction of HCl with F⁻ in the second since they are going to change the number of moles present in the buffer after reaction.

    mol HF initial = 1.06 mol/L x 0.48 mol/L = 0.51 mol

    mol HF reacted with NaOH = 0.16 mol

    mol HF final = 0.51 mol - 0.16 = 0.35 mol

    mol F⁻ initial = 1.06 L x 1.12 mol/L = 1.19 mol

    mol F⁻ final = 1.19 mol +.16 mol = 1.35 mol

    pH = 3.14 + log (1.35 / 0.35) = 3.73

    For the addition of HCl we do the same calculations for the reaction of HCl with F⁻:

    mol reacted F⁻ = 0.32 mol

    mol F⁻ initial = 1.06 L x 1.12 mol/L = 1.19 mol

    mol F⁻ final = 1.19 mol - 0.32 0 = 0.87 mol

    mol HF final = 0.51 mol + 0.32 mol = 0.83

    pH = 3.14 + log (0.87 / 0.83) = 3.16
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