[arq6]

Question: What is the pH of 0.5M NH4HCO3 solution?

Attempt 1: Added the two equations

1) NH4+ <=> NH3 + H⁺
2) H2CO3 <=> HCO3- + H⁺

Let x = [H+], Ka1 = Ka of NH4+ = 5.62 * 10^(-10) , Ka2 = Ka of H2CO3 = 1.7 * 10^(-4)
Combining the Ka equations 1) and 2),

(Ka1)*(Ka2) = [NH3][H+][H+][HCO3-]/[NH4+][H2CO3]

(Ka1)*(Ka2) = (x)(x)(x)(0.5-x) / (0.5-x)(x)

(Ka1)*(Ka2) = (x^2)

\sqrt{(Ka1)*(Ka2)} = x

Therefore, x = 3.09 * 10^(-7) M = [H+]
Therefore, pH = 6.509...

But I believe the answer is really around pH = 7.9

What am I doing incorrectly?

Ignore those empty bullets, if they appear in this post

[Arkcon]

I removed the brackets that caused the bullet points.  I hope I preserved the parts of your posting that you mean to say.  If I haven't you can try using LaTex, but that would be excessive for the amount of formulas you're writing.

[Borek]

Ka2 = Ka of H2CO3 = 1.7 * 10^(-4)

Doesn't look correct to me.

[Babcock_Hall]

My apologies if this confuses the issue, but I was under the impression that the effective pK_a of carbonic acid was higher than the value you provided.  The reason why it is higher is that there is an unfavorable equilibrum:  H₂O +CO₂  H₂CO₃.