Calculate the concentration of CH3COOH AND CH3COONa of buffer solution with pH 5.6.The total concentration of CH3COOH +CH3COONa is 0.2mol/mL.

Question

I need a quick asnwer

Answer 1

We need to use the Henderson Hasselbalch equation ...

pH = pKa + log(A-/HA)

A- is the conjugate base aka the acid without the proton (CH₃COO-)

HA is the acid with the proton (CH₃COOH)

The pKa = 4.76 (I just looked this up)

Plugging in what we know we get...

5.6 = 4.76 + log( [CH₃COO^-] / [CH₃COOH] )

This will let us solve for the ratio of the conjugate base : acid

You should get...

[CH₃COO^-]/[CH₃COOH] = 6.9

Since the two concentrations total 0.2 M, we can say...

[CH₃COO^-] + [CH₃COOH] = 0.2

Rearrange and get...

[CH₃COO^-] = 0.2 - [CH₃COOH]

Use that value and plug it into the equation above...

[CH₃COO^-]/[CH₃COOH] = 6.9

(0.2 - [CH₃COOH] )/( [CH₃COOH] ) = 6.9

Rearrange and solve for [CH₃COOH] and get...

[CH₃COOH] = 0.025 M

which means [CH₃COO-] = 0.2 - 0.025

[CH₃COO^-] = 0.175 M

Answer 2

You need to use the Henderson-Hasselbalch equation: 

pH = PKa + log [CH3COONa] / CH3COOH]

You know the pH and can look up the pKa of acetic acid. The salt and acid concentrations can be expressed as

[CH3COOH] = 0.2 – [CH3COONa]. 

The ratio needed for the log can be written [CH3COONa] / (0.2 – [CH3COONa]).

Solve for the log ratio for the only unknown [CH3COONa], then get [CH3COOH] by 0.2 – [CH3COONa]