Question

Calculate pH of buffer + HCl
Calculate pH of buffer + NaOH
(using attached calculations)
Given the following information:
Half of buffer solution (32mL) poured into a 150mL beaker. Label two beakers 1 and 2.
- 1.0 mL of 6.0M HCl added to beaker 1
- 1.0 mL of 6.0M NaOH added to beaker 2

Answer 1

Since you're dealing with a buffer, you can simplify this by using the henderson-hasselbach equation: \(\sf \color{red}{pH = pK_a + log[\frac{product}{reactant}]}\)

Answer 2

Just to add: the total amount of buffer solution that was prepared was 66.4mL (8.8mL of 3.0M acetic acid and 55.6 mL D.I. water)

Answer 3

Thank you @abb0t. For calculating pH of buffer + HCl how would i set that up?

Answer 4

Find the concentration of the product and reactant upon addition of both acid or base.

It might help if you find the equivalence point pH by using M\(_1\)V\(_1\)=M\(_2\)V\(_2\). This will help you see the pH should be below that.

Answer 5

Hm sorry I'm a little confused :/

Answer 6

Nevermind, just ignore that, you might not have enough information anyways.

Answer 7

I think you're missing information here. I don't know the concentration initially.

Answer 8

nor the Pka.

Answer 9

the pKa can be looked up pKa = 4.75 (acetic acid)

As abb0t mentioned, you wanna find the actual concentration of Acetic acid after the dilution:

\(8.8mL*3.0M=M_2*66.4mL\)

\(M_2=\dfrac{8.8mL*3.0M}{66.4mL}=0.3975903~M~\approx0.4~M\)

Now we need to find the concentrations of each species, acetic acid and acetate

\(K_A=10^{-4.75}=\dfrac{[AcO^-][H^+]}{[AcOH]}\)

I'm gonna skip the ICE table because you know how to do that

\(K_A=10^{-4.75}=\dfrac{x^2}{0.4~M-x}\)


Now we use the Henderson Hasselbalch

\(pH=pKa+log\dfrac{[AcO^-]}{[AcOH]}\)

Now we need the moles of HCl or the concentration in the whole volume.

Use: \(M_1V_1=M_2V_2\)

\(M_{HCl}=\dfrac{1mL*6~M}{33~mL}=0.1818~M\approx 0.182~M\)

Because were adding acid, were protonating some of the acetate and making some acetic acid.


\(pH=pKa+log\dfrac{[AcO^-]-[HCl]}{[AcOH]+[HCl]}\)

It seems like you're going to protonate all acetate. You have to account for this, i'm sure you can figure it out