need help 1. Assume that 50.0 mL of 0.100 M HCl Ka for acetic acid is 1.75 x 10

Question

need help

1. Assume that 50.0 mL of 0.100 M HCl Ka for acetic acid is 1.75 x 10 ·was added to 125 mL of 0.100 M Sodium Aceto. The solution. The 125 m acid is 175 er? Why o y not? Calculate the pH of the resulting so 2. Suppose you wanted to make a pH = 4.25 buffer. What would be the best acid/base system acetic acid/acetate or lactic acid/lactate? The Ka of lactic acid is 8.32 x 104. Thoroughly explain your reasoning. 3. Describe 3 different ways you could make an acetic acid / sodium acetate, deionized hydroxide acetate buffer with acetic acid, water, a solution of hydrochloric acid, and a solution of sodium

Explanation / Answer

Q1

mmol of acid = MV = 0.1*50 = 5

mmol of acetate = MV = 0.1*125 = 12.5

after reaction

acetic acid formed =5

acetate left = 12.5-5 = 7.5

therefore, since there is acetate ion (conjugate base) + acetic acid

this msut be ab uffer!

pH = pKa + log(acetate/acetic acid)

pH = 4.75+log(7.5/5)

pH = 4.926

Q2

if pH goal is 4.25, then

compare with pKa of lactic acid

pKa = -log(Ka) = -log(8.32*10^-4) = 3.08

nearest is pKa for acetica cid, 4.75

units of difference acetate = 4.75-4.25 = 0.5

units of difference, lactate= 4.25-3.08 = 1.17

Q3

i)

add acetic acid + sodium acetate --> resulting acetic acid + acetate in solution

ii)

add sodium acetate, and add some HCl for creation of acetic acid...in solution lefT = acetic acid + acetate

iii)

Add acetic acid, then add some base, NaOH.

Acetate will be formed, and someacetic aicd will be left

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