Question

Determine the pH of the solution of 100 mL of 0.150 M benzoic acid, HC7H5O2 (Ka=6.3x10-5 ). To this 25.0 mL of 0.400 M KOH is added, calculate the pH of the resulting solution. In a second step an additional 12.5ml of 0.400M KOH is added, calculate the pH of the resulting solution.

Answer 1

HC₇H₅O₂ + H₂O <=> (C₇H₅O₂)^-1 + H₃⁺O

0,150 M

0,150 M-X x x

Ka= ([(C₇H₅O₂)^-1] ^. [H₃⁺O]) / [HC₇H₅O₂]

6.3x10-5 = x.x/ (0,150-x)

6.3x10-5 = x²/ (0,150-x) ; Notice that [HC₇H₅O₂]= 0,150-x we can eliminate x because it is despicable.

so

6.3x10-5 = x²/ (0,150) => (6.3x10-5).(0,150)=x²  => 9.45x10-6=x² => square root of 9.45x10-6= x

x= 3,07x10-3 , it means that

[(C₇H₅O₂)^-1]=[H₃⁺O]= 3,07x10-3 ------> pH= -log [H₃⁺O] ------> pH= -log [3,07x10-3] ------>

pH=2,51

First step

When 25.0 mL of 0.400 M KOH is added the numer of moles is

M= Moles/Volumen(Lts) ------> solving moles,   Moles= M.V   ------> Moles= (0.400 Mol/L).(0.025Lts)

------> Moles= 0.010 moles of KOH..

we have to do the same with the acid

100 mL of 0.150 M benzoic acid

M= Moles/Volumen(Lts) ------> solving moles,   Moles= M.V   ------> Moles= (0.100 Mol/L).(0.Lts)

------> Moles= 0.015 moles of benzoic acid.

pH= -logKa + Log ((Moles HC₇H₅O₂ - Moles KOH) / (moles (C₇H₅O₂)^-1 +moles KOH))

pH= -log6.3x10-5 + Log ((0.015 moles - 0.010 moles) / (3,07x10-3 moles +0.010 moles))

pH= 4.20-0.41

pH= 3.79

In the second step.

We have to do the same

When 12.5ml of 0.400M KOH is added the numer of moles is

M= Moles/Volumen(Lts) ------> solving moles,   Moles= M.V   ------> Moles= (0.400 Mol/L).(0.0125Lts)

------> Moles= 0.005 moles of KOH..

Moles of HC₇H₅O₂ are ---> 0.015 moles - 0.010 moles = 0.005 moles

and Moles (C₇H₅O₂)^-1 are ---> 3,07x10-3 moles +0.010 moles = 0.01307 moles

applying again the formule

pH= -logKa + Log ((Moles HC₇H₅O₂ - Moles KOH) / (moles (C₇H₅O₂)^-1 +moles KOH))

pH= -log6.3x10-5 + Log ((0.005 moles - 0.005 moles) / (0.01307 moles +0.005 moles))

pH= 4.20