Question

For problem s 1 and 2 , assume an ac tivity coefficient of 1 for all substances and no effect of ionic strength. Eliminate terms in quadratic solutions for [H + ] only if the weak acid is dissociated < 5%. Reported pKa values can vary depending on the conditions under which they were measured ; therefore, in solving the following problems use the pKa values given with the problems.

1c . What is the pH of 45 mM H3PO4 ? (For c - f: Phosphoric acid (H3PO4 ) is a triprotic acid; pKa 1 = 2.12, pKa 2 = 7.21, pKa 3 = 12.32 Acetic acid pKa = 4.75)

d . What is the ratio of H2PO4 - to H3PO4 at a pH of 3?

e . If 2 volumes of 18 mM KOH are mixed with 1 volume of 22.5 mM H 3 PO 4 , what will be the pH of the final mixture?

f . If 15 μmoles of acetic acid is generated in a 1.0 ml enzymatic reaction buffered by 50 mM Na - phosphate (pH 7.0), what will be the final pH of the reaction mixture? Would the change in pH be smaller or larger if the reaction were buffered by 50 mM Na phosphate ( pH 6.0 )? Explain you answer; no calculations required.

Answer 1

1c pH of 0.45 mM H3PO4

H3PO4 <==> H2PO4- + H+

let x amount has dissoicated,

Ka1 = [H2PO4-][H+]/[H3PO4]

pKa1 2.12 = -logKa1

Ka1 = 7.58 x 10^-3 = x^2/0.045

x = [H+] = 0.0185 M

pH = -log[H+] = 1.73

d. ratio of H2PO4 to H3PO4 at pH 3

pH = 3 = -log[H+]

[H+] = 1 x 10^-3 M

Ka1 = 7.58 x 10^-3 = [H2PO4-] x 1 x 10^-3/[H3PO4]

[H2PO4-]/[H3PO4] = 7.58

e. moles of H3PO4 = 0.0225 M x 0.010 L = 2.25 x 10^-4 mol

moles of KOH added = 0.018 M x 0.020 L = 3.6 x 10^-4 mols

moles of HPO4^2- = 1.1 x 10^-4 mols

moles of H2PO4- = 1.15 x 10^-4 mols

[HPO4^2-] = 1.1 x 10^-4/0.03 = 3.67 x 10^-3 M

[H2PO4-] = 1.15 x 10^-4/0.03 = 3.83 x 10^-3 M

pH = 7.21 + log(3.67 x 10^-3/3.83 x 10^-3) = 8.17

f. 15 umol Acetic acid

Ka = 1.78 x 10^-8 = x^2/15 x 10^-6

x = [H+] = 5.16 x 10^-7 M

pH = 6.29

At pH 6.0, more of acetic acid is formed and thus greater pH value will be observed.