AP CHEM
posted by Sigurd .
Hello, I'm working these homework questions out; however it is a series of questions and without one step correctly completed, it's nearly impossible to get the others. I posted my work and commented where I need assistance. Thank you so much in advance!!!
Potassium hydrogen phthalate is a weak solid acide w/ the formula KHc8H4O4 (often abbrv. KHP) and a formula weight of 204.23 g/ mole. It is often used to standardize a solution of a strong acid with an unknown concentration. When 1.328 g of KHp is dissolved in 50 mL of H20, phenolphthalein indicator is added, and then titrated w/ a solution of NaOH (conc unknown), it takes 42.85 mL to complete the titration. Answer the following questionsThe hydrogen phthalate ion HC8H4O4^ (a) acts as a weak acid in water.
a) write out full molecular/net ionic reaction btw KHP and the NaOH solution
NaOH(aq) + KHC8H4O4(aq) > H20 (l)+NaKC8H4O4 (aq)
Net.I.E. OH^ (aq) + HC8H4O4^ >H20 (l) + C8H4O4
b)Calculate the concentration of the NaOH solution
1.328 g KHP(1 mol KHP/204.23 g) (1 mol Naoh/ 1 mol KHP) = 0.006502 mols NaOH
M= 0.006502 mols/ 0.04285 L = 0.1517 M NaOH
c)It is found that after 16.50 Ml of the NaOH has been added that the pH is 4.972. Calculate the Ka of HC8H4O4.
pH = pka + log (A/HA)
% completion = 16.5 mL/42.85 mL = 0.3851= 38.51 %
4.972 = pka + log (38.51/61.49)
pka = 4.972log (38.51/61.49)
pka = 5.175
Ka = 10^5.175
Ka = 6.67 X 10 ^ 6
the next part (below) is where i'm having some trouble. I don't understand how the ratio works for the Kb is it still the same % completion or does it change???
d) calculate the Kb for the pthalate ion using the Ka from your answer above
pOH = 144.972
pOH = 9.028
pOH = pkb + log (HB+/B)
9.028 = pkb + log (what goes here??)

sorry for part c that should read 16.50 mL not MI

Potassium hydrogen phthalate is a weak solid acide w/ the formula KHc8H4O4 (often abbrv. KHP) and a formula weight of 204.23 g/ mole. It is often used to standardize a solution of a strong acid with an unknown concentration(Actually KHP is used to standardize a base and that is how it is being used here). When 1.328 g of KHp is dissolved in 50 mL of H20, phenolphthalein indicator is added, and then titrated w/ a solution of NaOH (conc unknown), it takes 42.85 mL to complete the titration. Answer the following questionsThe hydrogen phthalate ion HC8H4O4^ (a) acts as a weak acid in water.
a) write out full molecular/net ionic reaction btw KHP and the NaOH solution
NaOH(aq) + KHC8H4O4(aq) > H20 (l)+NaKC8H4O4 (aq)This is OK
Net.I.E. OH^ (aq) + HC8H4O4^ >H20 (l) + C8H4O4 This tartrate ion should have a double minus charge.
b)Calculate the concentration of the NaOH solution
1.328 g KHP(1 mol KHP/204.23 g) (1 mol Naoh/ 1 mol KHP) = 0.006502 mols NaOH
M= 0.006502 mols/ 0.04285 L = 0.1517 M NaOH
c)It is found that after 16.50 Ml of the NaOH has been added that the pH is 4.972. Calculate the Ka of HC8H4O4.
pH = pka + log (A/HA)
% completion = 16.5 mL/42.85 mL = 0.3851= 38.51 %
4.972 = pka + log (38.51/61.49)
pka = 4.972log (38.51/61.49)
pka = 5.175
Ka = 10^5.175
Ka = 6.67 X 10 ^ 6 I get this same number but since the pH in the problem is to three significant figures and all your other work is to four, I wonder if you shouldn't allow yourself another place.?
the next part (below) is where i'm having some trouble. I don't understand how the ratio works for the Kb is it still the same % completion or does it change???
d) calculate the Kb for the pthalate ion using the Ka from your answer above
pOH = 144.972
pOH = 9.028
pOH = pkb + log (HB+/B)
9.028 = pkb + log (what goes here??)
The easy way to to Kb is to remember that KaKb = Kw. You have Ka and Kw. OR you can remember that pKa + pKb = 14 and do it that way. You've done good work here. 
Where did you get the 61.49 from?
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