B)

Question

Please help me with these questions on this chapter.

A) What is the pH of a solution of .500M K2HPO4? Ka3= 4.2x10^-13
Aspirin (acetylsalicylic acid, C9H8O4) is a weak monoprotic acid. To determine its acid-dissociation constant, a student dissolved 2.00g of aspirin in 0.600L of water and measured the pH . What was the Ka value calculated by the student if the pH of the solution was 2.62? Express answer using 2 sig figs Please help me with these questions on this chapter.

A) What is the pH of a solution of .500M K2HPO4? Ka3= 4.2x10^-13
B) Aspirin (acetylsalicylic acid, C9H8O4) is a weak monoprotic acid. To determine its acid-dissociation constant, a student dissolved 2.00g of aspirin in 0.600L of water and measured the pH . What was the Ka value calculated by the student if the pH of the solution was 2.62? Express answer using 2 sig figs Please help me with these questions on this chapter.


B) Aspirin (acetylsalicylic acid, C9H8O4) is a weak monoprotic acid. To determine its acid-dissociation constant, a student dissolved 2.00g of aspirin in 0.600L of water and measured the pH . What was the Ka value calculated by the student if the pH of the solution was 2.62? Express answer using 2 sig figs B) Aspirin (acetylsalicylic acid, C9H8O4) is a weak monoprotic acid. To determine its acid-dissociation constant, a student dissolved 2.00g of aspirin in 0.600L of water and measured the pH . What was the Ka value calculated by the student if the pH of the solution was 2.62? Express answer using 2 sig figs a student dissolved 2.00g of aspirin in 0.600L of water and measured the pH . What was the Ka value calculated by the student if the pH of the solution was 2.62? Express answer using 2 sig figs a student dissolved 2.00g of aspirin in 0.600L of water and measured the pH . What was the Ka value calculated by the student if the pH of the solution was 2.62? Express answer using 2 sig figs

Explanation / Answer

As we can see K2HPO4 is polyprotic acid. And only Ka3 is given.

We will write the reaction for 3rd dissociation only.

HPO4(2-) +H2O ßàPO4(3-) + H3O(+).

Ka = [PO4(3-)]*[H3O(+)]/[HPO4(2-)] = 4.2×10^-13

We can say that [PO3(3-)] = [H3O(+)]. And [H3O+] is [H+].

Therefore, Ka = [H+]*[H+] / [HPO4(2-)] = 4.2×10^-13

pKa = -log(Ka) = log[HPO4(2-)] - log[H+]^2 = -log (4.2×10^-13)

So, log[H+]^2 = log[HPO4(2-)] + log (4.2×10^-13)

2log[H+] = log[HPO4(2-)] + log (4.2×10^-13)

Now, pH = -log[H+], but we have two [H+] therefore multiplying pH by 2.

2pH = - log (4.2×10^-13) - log[HPO4(2-)]

2pH = - log(4.2×10^-13) - log(0.550)

2pH = 12.6363
pH = 6.3181 = 6.32

B)

2.00 g C9H8O4 x (1 mole C9H8O4 / 180 g C9H8O4) = 0.0111 moles C9H8O4

0.0111 moles C9H8O4 / 0.600 L = 0.0185 M C9H8O4

Molarity . . . . .C9H8O4 + H2O <==> H3O+ + C9H7O4-
Initial . . . . . . . .0.0185 . . . . . . . . . . . .0 . . . . . .0
Change . . . . . . . .-x . . . . . . . . . . . . . .x . . . . . .x
At Equilib. . . . .0.0185 - x . . . . . . . . . . .x . . . . . .x

Ka = [H3O+][C9H7O4-] / [C9H8O4] = x^2 / (0.0185 - x)

pH = 2.62; [H3O+] = 10^-pH = 10^-2.62 = 0.00239 = x

Ka = (0.00239)^2 / (0.0185 - 0.00239) = 3.54 x 10^-4

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