Find the Ph of the solution.

Question

If 50 mL of 0.30 M propionic acid [PA] is added to 250 mL of 0.024 M NaOH, what is the resulting pH of the solution? pK = 4.87.

 

I keep getting an answer of 4.47, but I know that it should be 4.69

 

I use the Henderson Hasselbalch, correct?  This is what I am doing.

 

[HA] = (0.05)(0.3M)/0.3

[HA] = 0.05

 

[A-] = (0.25)(0.024M)/0.3

[A-] = 0.02

 

pH = pK+log [A-]/[HA]

pH = 4.87 + log 0.024/0.3

pH = 4.47

 

The TA said the answer is supposed to be 4.69

Answer 1

Hi Isaac,

 

Determine moles of propanoic acid, PA and NaOH before mixing:

 

0.25 L NaOH * 0.024 mol/L = 0.006 moles NaOH

0.05 L PA * 0.30 mol/L = 0.015 moles PA

 

write chemical reaction:

 

PA + NaOH --> NaP + H₂O

 

Determine moles of propanoic acid, PA and NaP, A- after mixing:

* Sodium propanoate is the conjugate base in this reaction.

 

Initial moles - reacted moles = moles PA left after mixing

0.015 - 0.006 = 0.009 moles PA

 

The moles of NaOH will be converted to NaP, the conjugate base.  

 

So we have;

 

Weak acid: AH = 0.009 mol

Conjugate Base: A - = 0.006 mol

 

pH = 4.87 + Log₁₀(0.006/0.009)

pH = 4.87 - 0.18

 

pH = 4.69