paraperiodic acid is weak acid with the formula H5IO6, and Ka=2.8x10^-2. For all
ID: 856180 • Letter: P
Question
paraperiodic acid is weak acid with the formula H5IO6, and Ka=2.8x10^-2. For all of the parts of this question it is acceptable to simplify the math by assuming the [conc.] -x=[conc.] or [conc.] +x=[conc.]. A.) Although paraperiodic acid is a polyprotic acid, only loss of one proton is significant. Write a reaction for the reaction of paraperiodic acid in water, assuming that it only gives up one H+ ion. B.) Calculate the pH of a solution that is initially 2.8 M in paraperiodic acid. C.) Calculate the pH of a solution that is 1.00 M in paraperiodic acid AND 0.958 M in sodium paraperiodate (NaH4IO6). D.) Calculate (or determine from other answers) the pH of a buffer containing 1.00 M paraperiodic acid and 0.956 M sodium paraperiodate (NaH4IO6).
Explanation / Answer
1. H5IO6 +H2O ---> H3O+ + H4IO6-
a2.8
2. Calculate the pH of a solution that is initially 2.8 M in paraperiodic acid
ka = [H4IO6-] [H3O+] / [ H5IO6]
at equilibrium
H5IO6 +H2O ---> H3O+ + H4IO6-
2.8 -x x x
ka = [x] [x] / [2.8-x]
2.8x10^-2 = x2 / 2.8-x
2.8-x = 2.8
so 2.8x10^-2 = x2 / 2.8
x2 = 0.0784
x = 0.28
so conc [H+] = 0.28
Therefore pH = -log 2.8 = 0.55
3.
Calculate the pH of a solution that is 1.00 M in paraperiodic acid AND 0.958 M in sodium paraperiodate (NaH4IO6)
pH = pKa + log [salt / acid]
pH = 1.55 + log [0.958 / 1]
pH = 1.55 -0.0186 = 1.531
4. the above solution is a buffer so the ph of buffer is 1.531
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