This is from the last in Lecture 17 2016 Butops. It pretty well captums what we

Question

This is from the last in Lecture 17 2016 Butops. It pretty well captums what we need to know in the equilibrium chapters An Intermediate used in the synthesis of perfumes is valeric acid, also called pentanole acid. The pK_a of pentanoic acid is 4.84 at 25 degree C What is the pH of a 0.259 M solution of pentanole acid? Sodium pentanoate is added to make a buffered solution. What is the pH of the solution if it is 0.210 M in sodium pentanoate? What is the final pH if 8.00 mL of 0.100 M HCL are added to 75.0 mL of the buffered solution? What is the final pH if 8.00 mL of 0.100 M NaOH are added to 75.0 mL of the buffered solution?

Explanation / Answer

Q1

M = 0.259 of acid

HA <->H+ + A-

Ka = [H+][A-]/[HA]

10^- 4.84 = (x*x)/(0.259-x)

x = 0.0019

pH = -logx = -log(0.0019)

pH = 2.721

Q2.

pH = pKa + log(A-/HA)

pH= 4.84 + log(0.21 / 0.259) = 4.74891

Q3.

afterV = 8 mL of acid M = 0.1 in V = 75 mL buffer

mmol of acid = MV = 8*0.1 = 0.8 mmol

mmol of conjugate = 0.21*75 = 15.75 mmol

then

conjugate reacted = 15.75- 0.8 = 14.95

so...

mmol of acid = MV = 0.259*75 = 19.425

pH = 4.84 + log(14.95/19.425 ) = 4.726

Q4

after V = 8 mL of base M = 0.1 in V = 75 mL buffer

mmol of base = MV = 8*0.1 = 0.8 mmol

mmol of acid = 0.21*75 = 15.75 + 0.8 =16.55mmol

then

conjugate reacted = 19.425- 0.8 = 18.625

so...

mmol of conjugate = MV = 0.259*75 = 19.425

pH = 4.84 + log(16.55/19.425 ) = 4.77043

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