Chemistry Problem Related to Acid-Base Buffers

I'm trying to study for an exam and came across these two problems that I got completely stuck on. Can someone please go through the process of solving these problems and show me the steps you used to get your answer?

1. What is the [H3O+] and the pH of a benzoic acid -benzoate buffer that consists of 0.11 M C6H5COOH and 0.29 M C6H5COONa? (Ka of benzoic acid= 6.3*10^-5

2. What is the component concentration ratio [ (NO2-)/(HNO2) ] of a buffer that has a pH of 3.89? (Ka of HNO2=7.1*10^-4)

Thanks again for any help and quickest correct answers get the ten points!

1. Let benzoic acid be called BzOH. Then BzOH ===> H+ + BzO-

Ka = [H+][BzO-]/[BzOH] = 6.3 x 10^-5

Let [H+] = x. Then [BzO-] = 0.29 + x and [BzOH] = 0.11 - x. But x is so small that we can neglect it and say [BzO-] = 0.29 and [BzOH] = 0.11.

Ka = (x)(0.29)/(0.11) = 6.3 x 10^-5

x = 2.4 x 10^-5 M = [H+]

pH = -Log[H+], and Log(2.4x10^-5) = -4.6, so pH = +4.6 to two significant figures

2. HNO2 = H+ + NO2-

Ka = [H+][NO2-]/[HNO2] = 7.1 x 10^-4

pH = 3.89, so Log[H+] = -3.89, and Antilog(-3.89) = 1.29 x 10^-4 = [H+]

Ka = (1.29x10^-4)[NO2-]/[HNO2] = 7.1 x 10^-4

[NO2-]/[HNO2] = 5.5/1.0

1. So first, you have to find the pH because it would be easier. You should use the henderson-hasselbalch equation to make the finding of the pH easier. the henderson-hasselbalch equation is:

pH = pKa + log [conjugate acid]/[acid]
= -log(6.3x10^-5) + log (0.29)/(0.11)
= 4.20 + 0.42
pH = 4.62

now that you got the pH, you can compute for the [H3O+] from it.

pH = -log[H3O+]
[H3O+] = antilog-pH (antilog of the negative pH)
[H3O+] = 2.4 x 10^-5

2. Still you use the henderson-hasselbalch equation but now instead of looking for the pH, you are looking for the ratio.

pH = pKa + log [conjugate acid]/[acid]
log [conjugate acid]/[acid] = pH - pKa = 3.89 - (-log(7.1x10^-4)) = 3.89 - 3.15 = 0.74
[conjugate acid]/[acid] = antilog(0.74)
= 5.51 part NO2- per 1 part HNO2

So the ratio between NO2-/HNO2 is 5.51 per 1. :)