Question

A buffer solution is prepared by mixing 20.0 mL 0.45 M HAc (acetic acid) with 35.0 mL 0.45 MNaAc (sodium acetate) (a) What is the amount of 4.0 M HAc which must be added to this buffer solution to double [H3O+]? (b) What is the amount of 2.0 M 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)=

in order to double [H3O+]

this means

initial pH:

pH = pKa + log(A-/HA)

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

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

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

so..

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

to double..

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

so..

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

so..

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)

0.8729 *  (9 + HA) =  (5.75 - HA)

0.8729 *9 + 0.8729 HA = 5.75 - HA

1.8729 HA = 5.75 - 0.8729 *9

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+

find H+

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

5.4954*(9) - 15.75 = -6.4954H

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