Question

A buffer solution is prepared by mixing 20.0 mL 0.45M HAc (acetic acid) with 35.0 mL 0.45M NaAc (sodium acetate)

(a) What is the amount of 4.0M HAc which must be added to this buffer solution to double [H3O+]?

(b) What is the amount of 2.0M HCl that must be added to decrease the pH by 0.50?

(c) How much NaOH(s) in g has to be added to the solution to raise the pH by 2.00?

Answer 1

(a)

pH = pKa + log(A-/HA)

mmol of A- = 35*0.45 = 15.75 mmol of A-

mmol of HA = 20*0.45 = 9 mmol of HA

pH = 4.75 + log(15.75/9) = 4.9930

[H3O+] = 10^-pH = 10^-4.9930 = 1.016248*10^-5

to double..

2[H3O+] = 2*1.016248*10^-5 = 2.03249*10^-5

pH = -log(X) = -log( 2.03249*10^-5) = 4.691

pH = pKa + log(A-/HA)

4.691 = 4.75 + log(15.75 - HA / 9 + HA)

10^(4.691 -4.75 ) = (5.75 - HA) / (9 + HA)

0.8729 = (5.75 - HA) / (9 + HA)

HA = (5.75 - 0.8729 *9 ) / 1.8729   

HA= 1.1245 mmol required

MV = 1.1245

V = 1.1245/M = 1.1245/0.45 = 2.498 mL required

b)

HCL for... pH decrease of 0.5

new pH:

4.9930+0.5 = 5.49

so

5.49 = 4.75 + log(A-/HA)

10^(5.49-4.75) = A-/HA

5.4954*(HA) = A-

5.4954*(9 + H+) = 15.75 - H+

H+ = (5.4954*(9) - 15.75 ) / (-6.4954) = 5.1896

mmol of H+ = 5.1896

mmol = MV

5.1896 = 2*v

V = 5.1896/2 = 2.5948 mL of HCl

c)

6.99= 4.75 + log(A-/HA)

10^(6.99-4.75) = A-/HA

173.780HA = A-

173.780(9 - mmol of NaOH) = 15.75 + mmol of NaOH

173.780(9) - 173.780 mmol of NaOH = 15.75 + mmol of NaOH

173.780(9) - 15.75 = 174.780 mmol of NaOH

mmol o fNaOH = (173.780(9) - 15.75 )/174.780 = 8.8583 mmol = 8.8583*10^-3 mol

mass = mol*MW = 8.8583*40*10^-3 = 0.354332 g of NaOh