Question

1. Given: K. for CH,COOH - 1.8x106 50 mL of 0.1 M CH,COOH is titrated with 0.1 M NaOH (a) Write down the balanced neutralization equation: Answer: - (ii) (iii) (iv) no. of mole of CH,COOH How many mole of NaOH is needed to react with the no. of mole of CH,COOH found in (11)? From your answer in (ili), find the volume of NaOH needed to reach the equivalent point (b) What is the initial pH? (before the addition of any NaOH) Answer: (c) What is the pH when 25 mL of 0.1 M NaOH is added ? Note: there is a short-cur way to do this question if you remember the rule from the online lecture video Answer: (d) What is the pH when 50 mL of 0.1 M NaOH is added? Answer: (c) What is the pH when 70 mL of 0.1 M NaOH is added? Answer:

Answer 1

a) Balanced Neutralization reaction : CH3COOH (aq) + NaOH (aq) $\rightarrow$ CH3COONa (aq) + H2O (l)

ii) We have relation, molarity = No. of moles of solute / Volume of solution in L

$\therefore$ No. of moles of solute = Molarity $\times$ Volume of solution in L

No. of moles of solute = 0.1 mol / L $\times$ 0.050 L = 0.0050 mol

iii ) According to reaction, 1 mol NaOH required to react with 1 mol CH3COOH. Hence, moles of NaOH required to react with 0.0050 mol CH3COOH would be

0.0050 mol CH3COOH $\times$ ( 1 mol NaOH / 1 mol CH3COOH ) = 0.0050 mol NaOH

iv) We have relation, molarity = No. of moles of solute / Volume of solution in L

$\therefore$ Volume of solution in L = No. of moles of solute / molarity = 0.0050 mol / 0.1 mol / L = 0.050 L = 50 ml

Volume of NaOH required to reach the equivalence point = 50 ml

b) Initial pH of solution

a) pH of solution after 0.00 ml NaOH added.

Before addition of NaOH , solution contain CH3COOH only. The pH of solution will be due to dissociation CH3COOH in water.

Consider a dissociation of CH3COOH in water.

CH3COOH (aq) + H2O (l)$1585261151132_blob.png$CH3COO -(aq) + H3O +(aq)  

For above  reaction, Ka = [CH3COO - ] [H3O + ] / [CH3COOH ] = 1.8 $1585261151158_blob.png$ 10 -05

Consider X moles of acetic acid dissociated at equilibrium, then we can write

Ka = X $1585261151161_blob.png$ X / 0.1 - X = 1.8 $1585261151193_blob.png$ 10 -05

Acid is a weak acid. Hence, we can assume  X is very small as compared to 0.200. Hence we can write 0.1 - X $1585261151179_blob.png$ 0.1

Therefore, X 2 / 0.1 = 1.8 $1585261151197_blob.png$ 10 -05

  X 2 = 0.1 $1585261151221_blob.png$ 1.8 $1585261151242_blob.png$ 10 -05

X 2 = 1.8 $1585261151273_blob.png$ 10 -06

Taking square root on both sides, we get

X =1.34 $1585261151293_blob.png$ 10 -03 M = [CH3COO - ] =[H3O + ]

We have relation , pH = -log [H3O + ]

Therefore, pH = - log 1.34 $1585261151292_blob.png$ 10 -03  = 2.87

ANSWER : Initial pH = 2.87

C) pH when 25 ml 0.1 M NaOH added.

We know that, Volume of NaOH required to reach the equivalence point = 50 ml

$\therefore$  Volume of NaOH required to reach the half equivalence point = 25 ml

We know that, pH corresponding to half equivalence point = pKa of acid.

We have relation, pKa = - log Ka

$\therefore$ pKa = - log 1.8 $1585261151193_blob.png$ 10 -05

pKa = 4.74

d) pH when 50 ml of 0.1 M NaOH added

At equivalence point, all acetic acid is consumed by added NaOH. pH of solution will be due to dissociation of CH3COONa in water.

Consider hydrolysis of C6H5COONa.

CH3COO- (aq) + H2O (l)$1585261710157_blob.png$ CH3COOH (aq) + OH -(aq)

For above reaction, Kb = [CH3COOH] [OH - ] / [CH3COO- ]= Kw / Ka = 10 -14 /  1.8 $1585261710086_blob.png$ 10 -05 = 5.56 $1585261710153_blob.png$ 10 -10

Moles of CH3COONa.= Moles of NaOH added = 0.0050 mol

Volume of solution= volume of acid + volume of base = 50.0 + 50.0 =100.0 ml = 0.100 L

[CH3COONa.] = 0.0050 mol / 0.100 L =0.05 M

Consider X moles of sodium acetate dissociated at equilibrium, then we can write

Kb = X $1585261710165_blob.png$ X / 0.05 - X =5.56 $1585261710170_blob.png$ 10 -10

Here X is very small as compared to 0.050. Hence we can write 0.050 - X $1585261876529_blob.png$ 0.050

X 2 = 0.050 $1585261876625_blob.png$ 5.56 $1585261876613_blob.png$ 10 -10

= 2.78 $1585261876607_blob.png$ 10 -11

X =5.27 $1585261876646_blob.png$ 10 -06 M = [OH -]

We have pOH = - log [OH -] = - log5.27$1585261876677_blob.png$ 10 -06 = 5.28

We have, pH + pOH = 14

Therefore, pH = 14 -5.28 = 8.72

ANSWER : pH = 8.72

e) pH after addition of 70 ml of 0.1 M NaOH

After equivalence point, pH of solution will be due to excess NaOH in the solution.

mmol of NaOH added = 0.1 $\times$ 70 = 7 mmol

mmol of CH3COOH = 0.1 $\times$ 50 =5 mmol

Excess mmol of NaOH = 7 - 5 = 2 mmol

Volume of solution = 50 + 70 = 120 ml

[NaOH] = 2 mmol / 120 ml = 0.0167 M

We have relation , pOH = -log [OH-] = - log 0.0167 = 1.79

pH = 14 -1.79= 12.21