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**1. The problem statement, all variables and given/known data**

A vessel contains 500mL of .1 molar H2S solution. For H2S Ka1 = 1 x 10^-7

Ka2 = 1.3 x 10^-13.

What will the pH be when 800mL of .1molar NaOH has been added.

**2. Relevant equations**

ph = -log([H])

ph = pka + log(B/A)

**3. The attempt at a solution**

This is how I started, I found the moles of each. I found I started with .08 moles of NaOH and .05 moles of H2S, subtracted them to get .03 moles of NaOH. Then .03 / (.5 + .8) to get the concentration (total volume), then because all thats left is strong base I took the -log(.023) and then subtracted that from 14 to get the pH. I keep ending up with 12.36 but the answer is 13.1. I need to be able to do this for my exam tomarrow, can anyone help?

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