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1. 25.0mL of 0.20M Propanic acid (HC3H5O2, Ka = 1.3×10-5) is titrated using 0.10M NaOH. Calculate the following pH. Show your work.

a. When 0.0 mL NaOH is added.

b. When 25.0 mL NaOH is added

c. When 50.0 mL NaOH is added

d. When 75.0mL NaOH is added.

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1 answer
  1. Let's call the acid HP.
    a.
    .......HP ==> H^+ + P^-
    I.....0.2M....0.....0
    C......-x.....x.....x
    E....0.2-x....x.....x

    Plug the E line into the Ka expression and solve for x = (H^+), then convert to pH.

    b is done with the Henderson-Hasselbalch equation.

    c is the equivalence point and the rxn is HP + NaOH ==> NaP + H2O
    The pH is determined by the hydrolysis of the salt NaP. The concentration of the salt is mols NaP/L. mols NaP = M x L = 0.2 x 0.025 = 0.005. L = 50 mL from the beginning + 50 mL added NaOH = 75 mL = 0.075 L so M NaP = 0.005/0.075 - 0.0667. The hydrolysis works this way.
    .........P^- + HOH ==> HP + OH^-
    I.....0.0667...........0.....0
    C......-x..............x.....x
    E....0.0667-x..........x.....x

    Kb for P^- = (Kw/Ka for HP)
    = (x)(x)/(0.0667-x). Solve for x = (OH^-) and convert to pH.

    D. You have excess OH^-. Calculate that and convert to pH.

    Post your work if you have questions and PLEASE tell us exactly what you don't understand.

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