Question

1. Using a 0.20 M phosphate buffer with a pH of 6.9, you add 0.80 mL of 0.46 M HCl to 53 mL of the buffer. What is the new pH of the solution? (Enter your answer to three significant figures.)

2. Using a 0.20 M phosphate buffer with a pH of 6.9, you add 0.80 mL of 0.46 M NaOH to 53 mL of the buffer. What is the new pH of the solution? (Enter your answer to three significant figures.)

Answer 1

1.

For phosphoric acid, pKa1 = 2.15,

pKa2 = 7.20,

pKa3 = 12.3

Here pH = 6.9;

Molarity of buffer = 0.20 M

Volume of buffer = 53 ml

Molarity of HCl = 0.46 M

Volume of HCl= 0.80 ml
means HPO4^2-/H2PO4- buffer, with pKa2 = 7.2

6.9 = 7.2 + log [HPO42-]/ [H2PO4-]
-0.3 = log [HPO42-]/ [H2PO4-]

[HPO42-]/ [H2PO4-] = 10^-0.3

[HPO42-]/ [H2PO4-] =0.50

[HPO42-] = [H2PO4-] * 0.50   ---1

And total mole of [HPO42-]+ [H2PO4-] = molarity * volume in L

= 0.20*53/1000

[HPO42-]+ [H2PO4-] = 0.0106   ----2 from equation 1 and2;

[H2PO4-] * 0.50   + [H2PO4-] = 0.0106  

1.50[H2PO4-] = 0.0106

[H2PO4-] = 0.0071

Then ; [HPO42-] =   0.0035

moles H+ added = 0.46 x 0.80/1000

=0.000368 mole HCl or H++

HPO42- + H+ = H2PO4-
moles HPO42- = 0.0035 - 0.000368=0.003132

moles H2PO4- = 0.0071 + 0.000368=0.007468
total volume = 53 mL++ 0.80 ml = 53.8 m L

= 0.0538 L

[HPO42-]= 0.003132/ 0.0538= 0.058 M
[H2PO42-]= 0.007468 / 0.0538 L=0.139 M

pH = 7.2 + log [HPO42-]/ [H2PO4-]


pH = 7.2 + log 0.058 /0.139

=7.2- 0.38

= 6.82

2.

[H2PO4-] = 0.0071

Then ; [HPO42-] =   0.0035

moles OH- added = 0.46 x 0.80/1000

=0.000368 mole NaOH or OH-

H2PO4- + OH- = HPO42-

+moles HPO42- = 0.0035 + 0.000368 =0.003868
moles H2PO4- = 0.0071 - 0.000368=0.006732

total volume = 53 mL+ 0.80 ml = 53.8 m L

= 0.0538 L

[HPO42-]= 0.003868/ 0.0538= 0.072 M
[H2PO4-]= 0.006732 / 0.0538 L=0.125 M

pH = 7.2 + log [HPO42-]/ [H2PO4-]

pH = 7.2 + log 0.072/0.125

=7.2- 0.24

= 6.96