The triprotic acid H3A has ionization constants of Ka1 = 6.8× 10–3, Ka2 = 8.1× 10–9, and Ka3 = 5.0× 10–12. Calculate the following values for a 0.0760 M solution of NaH2A

[H+] = ?
[H2A-]/[H3A]= ?
Calculate the following values for a 0.0760 M solution of Na2HA.
[H+]= ?
[HA2-]/[H2A-]= ?

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To solve this problem, we need to consider the ionization reactions of the triprotic acid H3A and the corresponding dissociation constants.

For a 0.0760 M solution of NaH2A:
1. Determine the concentration of H3A:
NaH2A → Na+ + H2A-
Since NaH2A is dissociated into Na+ and H2A-, the concentration of H3A is equal to the concentration of NaH2A, which is 0.0760 M.

2. Calculate the concentrations of [H+] and [H2A-]/[H3A]:
Since H3A is triprotic, it undergoes three ionization reactions:
H3A ⇌ H+ + H2A- (Ka1)
H2A- ⇌ H+ + HA2- (Ka2)
HA2- ⇌ H+ + A3- (Ka3)

a) To determine [H+]:
We will consider Ka1 because H3A is dissociated to H+ and H2A- in the first step.
Ka1 = [H+][H2A-] / [H3A]
[H+] = Ka1 * [H3A] / [H2A-]

Substitute the given values:
[H+] = (6.8 × 10^(-3)) * (0.0760 M) / (0.0760 M)
[H+] = 6.8 × 10^(-3) M

b) To determine [H2A-]/[H3A]:
We will consider Ka1 because H3A is dissociated to H+ and H2A- in the first step.
[H2A-] / [H3A] = (Ka1 * [H3A]) / [H2A-]
[H2A-] / [H3A] = (6.8 × 10^(-3)) * (0.0760 M) / (0.0760 M)
[H2A-] / [H3A] = 6.8 × 10^(-3)

Therefore, for a 0.0760 M solution of NaH2A:
[H+] = 6.8 × 10^-3 M
[H2A-]/[H3A] = 6.8 × 10^-3

Now, let's calculate the values for a 0.0760 M solution of Na2HA:
1. Determine the concentration of HA2-:
Na2HA → 2Na+ + HA2-
Since Na2HA is dissociated into 2Na+ and HA2-, the concentration of HA2- is half the concentration of Na2HA, which is
(0.0760 M)/2 = 0.0380 M.

2. Calculate the concentrations of [H+] and [HA2-]/[H2A-]:
Now, we will consider Ka2 because HA2- is formed in the second ionization step.
Ka2 = [H+][HA2-]/[H2A-]
[H+] = (Ka2 * [H2A-]) / [HA2-]
[H+] = (8.1 × 10^(-9)) * (0.0380 M) / (0.0760 M)
[H+] = 4.05 × 10^(-9) M

[HA2-]/[H2A-] = (Ka2 * [H2A-]) / [H+]
[HA2-]/[H2A-] = (8.1 × 10^(-9)) * (0.0380 M) / (4.05 × 10^(-9) M)
[HA2-]/[H2A-] = 7.61

Therefore, for a 0.0760 M solution of Na2HA:
[H+] = 4.05 × 10^-9 M
[HA2-]/[H2A-] = 7.61

Okay, I think you can do it this way, but I am not sure:

B- + H20 ---> HB + OH

Kb=[OH-][HB-]/B

Kw=Ka*Kb=1 x 10^-14

Solve for kb=Kw/Ka= 1 x 10^-14/6.8× 10^-3=Kb

sqrt*(Kb*0.0760 M)=OH

-log(OH)=pOH

14-pOH=pH

10^(-pH)=H+ concentration.

Use the Henderson Hasselbalch equation for the second part. You know the pH and you know the pka=-log(ka)

pH=pKa=log(A-/HA), solve for the ratio.

Repeat the above steps for your second problem.