Calculate the pH after 0.13 mol of NaOH is added to 1.00 L of 0.60 M HF and 1.04 M KF?

Question

I'm not very good with buffer questions but I thought I was doing this correctly when I solved for the pH but I continually find the wrong answer.

I've set up

HF + OH --> KF

.6 (HF) .13 (OH) 1.04 (KF)
- .13 (HF) -.13 (OH) +.13 (KF)
_________________________
.47 (HF) 0 (OH) 1.17 (KF)

.47x / 1.17 = 7.2 * 10^-4(Ka) and solve for x and then take the -log of x to find the pH.

Apparently something is incorrect but I can't figure out what. Any help would be appreciated.

Answer 1

[H+] = Ka x 0.47/1.17.

I think that you have got the moles the wrong way round.

Answer 2

I would say but not sue that NaOH and HF are strong electrolytes. So KF is a neural salt .
The 0.13 mole of NaOH reacts with 0.13 moles of HF

So ,you have a solution of .47 HF

This HF should be completely ionized and yield a concentration of H+ 0.47

So pH = log (1/0.47) =0.33