Question

someone prepared 50ml of buffer A using .5moles of HA and .5 moles of A- while another person prepared 50ml of buffer B using .25moles of HA and .25moles of A-

a. do the two buffer solutions have same or different ph? b. if 1ml of .01M NaOH were added to the two buffer solutions, would the PH of the two solutions increase or decrease and why? c. which solution buffer A or B would show a smaller change in the pH for question b and why

Answer 1

(a)

we know that pH = pka+log ([A^-]/[HA])

here, for first part,

pH = pka+log ([0.5]/[0.5])= pKa

for second part:

pH = pka+log ([0.25]/[0.25]) = pKa

these is no chnage in pH .Both have same values.

(b)

when base NaOH of 10^-5 mole is added.

pH = pka+log ([0.5+10^-5]/[0.5-10^-5])= pKa +1.73 x10^-5

for case -2:

pH   = pka+log ([0.25+10^-5]/[0.25-10^-5])= pKa +3.47 x10^-5

In both cases, pH has been increased

(C)

For Buffer-B the increase is higher than buffer-A . This is because the ratio of A- to HA is higher in buffer-B than buffer-A