acid and base calculation question help

[allthebest]

Please help me with this question:

Determine the pH of a solution formed from adding 50.0cm3 of 1.00moldm-3 ethanoic acid, CH3COOH(aq), to 50.0 cm3 of 0.600moldm-3 sodium hydroxide, NaOH (aq)

mark scheme:

(initial)[CH3COOH] = 0.500 mol dm–3 and) eqm [CH3COOH] = 0.200 mol dm–3;
(initial)[CH3COO–] = 0.300 mol dm–3 and) eqm [CH3COO–] = 0.300 mol dm–3;
Allow 0.02 moles and 0.03 moles instead of 0.200 and 0.300 mol dm–3.

             [H+] = Ka  = 1.16 × 10–5 mol dm–3 / pH = pKa + log;
pH = 4.94;
Award [3 max] for correct final answer if no working shown.

I'm not sure where that "(initial)[CH3COOH] = 0.500 mol dm–3 and) eqm [CH3COOH] = 0.200 mol dm-3" came from...

any explanations?

and this questions is where you add ethanoic acid to NaOH, but why are there no calculations regarding NaOH??

I'm having difficulties with acid and base calculations... an explanation would be helpful.

Thank you.

[Msj Chem]

The reaction is:

CH₃COOH + NaOH ==> NaCH₃COO + H₂O

Use n=CV to calculate amount in mol of each.

CH₃COOH:

n=0.05 mol 

NaOH

n=0.03 mol

Now make an ICE box:

                 CH₃COOH + NaOH ==> NaCH₃COO + H₂O

initial            0.05           0.03                0               0

eqm             0.02              0              0.03            0.03

This gives 0.02 mol of CH₃COOH and 0.03 mol of CH₃COO^-

If you use C=n/V you get 0.200 moldm^-3 and 0.300 moldm^-3 

Then use the Henderson-Hasselbach equation to calculate the pH.

But be aware that the use of this equation is only necessary in option B and D (not topic 18). 

Edited November 28, 2016 by Msj Chem

[kw0573]

Since more acid is added than base, all of the NaOH neutralizes ethanoic acid. We get instead the same amount of the conjugate base, CH₃COO^- and the amount of unreacted CH₃COOH remaining. 

The equation Cv = Cv in dilution describe the change in concentration from 50 cm³ to when you mix the two solutions to 100 cm³.

(1 mol • dm^-3 )(50 cm³ )= new concentration * 100 cm³.

New concentration of ethanoic acid is thus 0.5 mol dm^-3 (which makes sense that it should be half as dilute because you doubled the volume). Similarly, concentration of NaOH is 0.3 mol dm^-3 (which is half of 0.6 mol dm^-3).  Because of strong base-weak acid reaction between NaOH and CH₃COOH, 0.3 mol dm^-3 is the effective concentration of CH₃COOH consumed. That left with 0.3 mol dm^-3 of the conjugate base and (0.5 - 0.3 = ) 0.2 mol dm^-3 of CH₃COOH. Then apply Henderson-Hasselbalch equation to get the pH.