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

Question

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-]= ?

h · Accepted Answer

whattup mike

Devron · Answer

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.

Jake · Answer

Sup

Bob · Answer

If you aint licking door knobs I don't know what is wrong with Phil Swift