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 questions-The 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 = 4.972-log (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 = 14-4.972
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 questions-The 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 = 4.972-log (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 = 14-4.972
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?

To calculate the Kb for the phthalate ion (C8H4O4^-), we can use the relationship between Ka and Kb for a conjugate acid-base pair:

Kw = Ka * Kb

Since Kw (the ionization constant of water) is a known value (1.0 x 10^-14 at 25°C), and we have the Ka value for HC8H4O4 (6.67 x 10^-6), we can rearrange the equation to solve for Kb:

Kb = Kw / Ka

Substituting the values:

Kb = (1.0 x 10^-14) / (6.67 x 10^-6)
Kb ≈ 1.5 x 10^-9

So, the Kb for the phthalate ion is approximately 1.5 x 10^-9.

In the equation pOH = pkb + log (HB+/B), the ratio of [HB+] to [B] represents the concentration of the conjugate acid and base. In this case, the conjugate acid is HC8H4O4^- (phthalate ion) and the conjugate base is C8H4O4 (phthalate). The concentration ratio can be determined by using the percent completion of the reaction.

The percent completion of the reaction is given as 38.51%. This means that 38.51% of the phthalate ion (HC8H4O4^-) has been converted to its conjugate acid (HB+) and conjugate base (B).

To calculate the concentration ratio, you can assume that the initial concentration of HC8H4O4^- is 100% and subtract the percent completion from 100%. Then, divide the remaining percentage equally between HB+ and B.

For example, in this case, if the percent completion is 38.51%, then the remainder is 100% - 38.51% = 61.49%. Divide this by 2 to get the concentrations of HB+ and B. In this example, each will be 30.75%.

So, the concentration ratio for the equation pOH = pkb + log (HB+/B) would be:

pOH = pkb + log (30.75/30.75)

Therefore, the expression in the logarithm will be 1, resulting in:

pOH = pkb + 1
pOH - pkb = 1

From here, solving for pkb would give you the value needed to calculate Kb using the equation mentioned earlier.