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Chemistry, Equilibrium Constant Question

  • Thread starter jaguar7
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Haha --- nvm, found it. Sorry/Thanks... ^_^

Where does the 2.19 come from?

How many moles of NaOH would need to be added to 500.0 mL of 0.70 M HClO to create a buffer with a pH of 7.80?

Initially, we have 0.35 mol of HClO. We are going to add NaOH which will react with the weak acid to create the conjugate weak base.

HClO + OH- → H2O + ClO-


It is this weak base we’ll need to have for the buffer. First, let’s find the ratio of the weak base to the weak acid using the Henderson-Hasselbalch equation:

7.80 = -log(3.5*10^(-8) + log(base/acid)

base/acid = 2.19

Since we will be making base from the weak acid, we will react x moles of OH-. So after the reaction takes place, we have 0.35 – x moles of acid and x moles of base.

x / (0.35 - x) = 2.19

x = 0.240 moles
 
Last edited:

Answers and Replies

  • #2
Borek
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Won't hurt if you will explain what you were asking for and what you have found. I guess I can guess, not so sure about others.
 

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