# Chemistry

posted by .

Given a diprotic acid, H2A, with two ionization constants of Ka1=3.9E-2 and Ka2=5.9E-8.

Calculate the pH and molar concentration of each protonated form for a 0.135 M solution of:
(a) H2A
(b) NaHA
(c) Na2A

• Chemistry -

For H2A it follows the usual form of ionizsation as if you had a monoprotic acid.
..........H2A ==> H^+ + HA^-
I.......0.135M....0......0
C.........-x......x......x
E........0.135-x..x.......x

Ka1 = (H^+)(HA^-)/(H2A)
Substitute and solve for x = (H^+) = (HA^-) and convert to pH. (H2A) = 0.135-x. I strongly suspect you will need t use the quadratic equation to solve this because ka1 is relatively large.

For NaHA.
(H^+) = sqrt [(k2(HA^-)+Kw)/(1+(HA^-)/k1)] and convert to pH.

For Na2A, set up an ICE chart for the hydrolysis of A^2-.

• Chemistry -

I'm not getting the right answers for the pH, [H2A], and [HA^-] in part a (0.135 M solution for H2A)

For H2A, I did
3.9E-2=x^2/(.135-x)

x^2+.039x-.005265=0
x=.06725
[H^+]=[HA^-]=.06725 M

pH=-log(.06725)=1.17

Finding [H2A]=.135 M-.06725 M=.06775 M

Finding [A^2-]=Ka2*[HA^-]/[H^]=5.9E-8 M

## Similar Questions

1. ### Chemistry

Given a diprotic acid, H2A, with two ionization constants of Ka1 = 3.9× 10–2 and Ka2 = 5.9× 10–8. Calculate the pH and molar concentrations of each protonated form for a: (a)0.135 M solution of H2A pH= [H2A]= [HA]= [A2-]= (b)0.135 …
2. ### Chemistry

Given a diprotic acid, H2A, with two ionization constants of Ka1 = 4.5× 10–2 and Ka2 = 5.2× 10–7. Calculate the pH and molar concentration of each protonated form for a 0.133 M solution of NaHA. Calculate the pH and molar concentration …
3. ### chemistry

For the diprotic weak acid H2A Ka1=2.1x10^-5 and Ka2=5.5x10^-7. What is the pH of a 0.0800M solution of H2A?
4. ### Chemistry

For the diprotic weak acid H2A, Ka1 = 2.2 × 10-5 M and Ka2 = 7.8 × 10-7 M. What is the pH of a 0.0650 M solution of H2A?
5. ### Chemistry

For the diprotic weak acid H2A, Ka1 = 2.2 × 10-5 M and Ka2 = 7.8 × 10-7 M. What is the pH of a 0.0650 M solution of H2A?
6. ### Chemistry

For the diprotic weak acid H2A, Ka1 = 2.2 × 10-5 M and Ka2 = 7.8 × 10-7 M. What is the pH of a 0.0650 M solution of H2A?
7. ### Chemistry

For the diprotic weak acid H2A, Ka1 = 2.2 × 10-5 M and Ka2 = 7.8 × 10-7 M. What is the pH of a 0.0650 M solution of H2A?
8. ### Chemistry

For the diprotic weak acid H2A, Ka1 = 2.2 × 10-5 M and Ka2 = 7.8 × 10-7 M. What is the pH of a 0.0650 M solution of H2A?
9. ### UIC

Given a diprotic acid, H2A, with two ionization constants of Ka1 = 3.57× 10–4 and Ka2 = 3.14× 10–12, calculate the pH and molar concentrations of H2A, HA–, and A2– for each of the solutions below. (a) a 0.182 M solution of …
10. ### CHEMISTRY!!

For the diprotic weak acid H2A, Ka1 = 2.0 × 10^-6 and Ka2 = 6.7 × 10^-9. What is the pH of a 0.0800 M solution of H2A?

More Similar Questions