# buffers

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9.00mL of 1.10M CH3COOH and 1.00mL of 0.900M CH3COO- what is the pH?

• buffers -

Use the Henderson-Hasselbalch equation.

• buffers -

we havn't learnt that equation.

• buffers -

pH = pKa + log(base/acid)
pKa is pKa for CH3COOH
base is (CH3COO^-)
acid is (CH3COOH)

• buffers -

thanks. so for this question is the pH 3.70?

• buffers -

Or you can do it another way.
Write ionization of CH3COOH.

CH3COOH ==> H^+ + CH3COO^-

Now write the Ka expression.
Ka = (H^+)(CH3COO^-)/(CH3COOH)

Now plug in (CH3COO^-) from the problem.
Substitute (CH3COOH) from the problem.
Substitute Ka for CH3COOH. Try to find Ka in your text or notes but it's very close to 1.8 x 10^-5. Calculate (H^+) and convert that to pH.

• buffers -

I calculate 3.698 pH which rounds to 3.70. Yes. Very good.

• buffers -

thanks a lot :D

• buffers -

You're welcome.

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