Please help, specially for (c) and (d).?

Calculate the pH of a buffer consisting of 0.50 M HF and 0.45 M KF (a) before and (b) after the addition of 0.40 g of NaOH to 1.0 L of the buffer

(Ka of HF = 6.8E-4). For sake of comparison,

calculate the pH before and after adding the same amount of NaOH to (c) 1.0 L of pure water and (d) 1.0 L of 0.50 M HCl.

Discuss any differences in the magnitude of change in the pH between the three scenarios.

1 Answer

Relevance
  • 5 months ago

    HF <--> H+ + F-

    Ka = 6.8X10^-4 = [H+][F-]/[HF]

    (a) 6.8X10^-4 = [H+](0.45) / 0.50

    [H+] = 7.56X10^-4 M

    pH = 3.12

    (a) Initial moles HF = 0.50 mol

    Initial moles F- = 0.45 mol

    Moles NaOH added = 0.40 g / 40 g/mol = 0.01

    Added NaOH quantitatively neutralizes 0.01 mol HF forming an additional 0.01 mol F-. So, after addition, [HF] = 0.49 and [F-] = 0.46

    6.8X10^-4 = [H+] (0.46)/(0.49)

    [H+] = 7.24X10^-4

    pH = 3.14

    Adding 0.01 mol NaOH to 1 L water (which initially has a pH = 7.0) results in [OH-] = 0.01 M

    pOH = 2.0

    pH = 12.0

    Adding 0.01 mol NaOH to 0.50 M HCl decreases [H+] from 0.50 M to 0.49 M

    pH before addition = 0.301

    pH after addition = 0.309

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