Question

Question 2 (20pts): A 20 mL sample of 0.01 M propionic acid (CH3CH2COOH; Pka = 4.87) is titrated with 0.05 M NaOH. A) Write out the chemical reaction for this titration. B) Calculate the initial pH of the sample. C) Calculate the volume of NaOH required to reach the equivalence point. D) Calculate the pH of the solution at the equivalence point. E) Sketch a titration curve for this titration (pH versus volume NaOH added). Note the location of the equivalence point on the curve.

Answer 1

A) CH₃CH₂COOH _(aq) + NaOH _(aq)  CH₃CH₂COONa _(aq) + H₂O (l)

B) pH of solution before addition of any NaOH.

Initially there is only CH₃CH₂COOH present in the solution, hence pH of solution will be due to dissociation of CH₃CH₂COOH.

Consider a dissociation of CH₃CH₂COOH in water.

CH₃CH₂COOH _(aq) + H₂O _(l) CH₃CH₂COO ^-_(aq) + H₃O ⁺_(aq)  

For above reaction, dissociation constant is Ka = [CH₃CH₂COO ^- ] [H₃O ⁺ ] / [CH₃CH₂COOH ]

Let's use ICE table.

Concentration (M)	CH3CH2COOH	CH3CH2COO -	H3O +
Initial	0.01
Change	- X	+X	+X
Equilibrium	0.01 - X	X	X

Therefore, K a = ( X) (X) / 0.01 - X = 10 ^-4.87 = 1.349 $\times$ 10 ^-05

X ² / 0.01 - X = 1.349 $\times$ 10 ^-05  

CH₃CH₂COOH is a weak acid. Hence, we can assume X is very small as compared to 0.01 Hence we can write 0.01- X 0.01.

X ² / 0.01 = 1.349 $\times$ 10 ^-05  

X ² = 0.01 (1.349 $\times$ 10 ^-05  )

X ² = 1.349 $\times$ 10 ^-07

X = 3.673 $\times$ 10 ^-04 = [CH₃CH₂COO ^- ] = [H₃O ⁺ ]

We have, pH = - log [H₃O ⁺ ]

$\therefore$ pH = - log 3.673 $\times$ 10 ^-04 = 3.43

ANSWER : pH of solution before addition of any NaOH = 3.43

C) Calculation of volume of NaOH required to reach equivalence point

Consider reaction, CH₃CH₂COOH _(aq) + NaOH _(aq)  CH₃CH₂COONa _(aq) + H₂O (l)

We have, M _acid$\times$ V _acid = M _base$\times$ V _base

   V _base = M _acid$\times$ V _acid / M _base

V _base = 0.01 M $\times$ 20 ml / 0.05 M

V _base = 4.0 ml

Volume of NaOH required to reach equivalence point = 4.0 ml

D)  pH of solution at equivalence point ( 4.0 ml)

At equivalence point, all CH₃CH₂COOH is consumed by added NaOH. pH of solution will be due to dissociation of CH₃CH₂COOH in water.

CH₃CH₂COO^- (aq) + H₂O _(l) CH₃CH₂COOH _(aq) + OH ^-_(aq)

For above reaction, dissociation constant is Kb = [CH₃CH₂COOH] [OH ^- ] / [CH₃CH₂COO^- ]= Kw / Ka

K b = 10 ^-14 / 1.349 x 10 ^-05 = 7.413 x 10 ^-10

mmol of CH₃CH₂COONa produced = mmol of NaOH added to the solution = 0.05 $\times$ 4.0 = 0.2 mmol

Volume of the solution = volume of CH₃CH₂COOH + volume of NaOH = 20 ml + 4.0 = 24.0 ml

[CH₃CH₂COONa] = No. of moles of CH₃CH₂COONa / volume of solution in L

= [ 0.2 /1000] /[ 24.0 /1000]

=0.00833 M

Let's use ICE table.

Concentration (M)	CH3CH2COO -	CH3CH2COOH	OH -
Initial	0.00833
Change	- X	+X	+X
Equilibrium	0.00833 - X	X	X

Therefore, K b = (X) (X) / 0.00833 - X

K b = X ² / 0.00833 - X = 7.413 x 10 ^-10

Here X is very small as compared to 0.00833. Hence we can write 0.00833 - X 0.00833.

X ² / 0.00833 = 7.413 x 10 ^-10

X ² = 0.00833 ( 7.413 x 10 ^-10 )

X ² = 6.175 x 10 ^-12

X = 2.485 x 10 ^-06 M = [CH₃CH₂COOH] = [OH ^- ]

We have relation, [H ⁺] [OH ^- ] = 10 ^-14

$\therefore$ [H ⁺ ] = 10 ^-14 / [OH ^- ] = 10 ^-14 / 2.485 x 10 ^-06 =4.024 x 10 ^-09 M

We have, pH = - log  [H ⁺ ] = - log 4.024 x 10 ^-09 = 8.40

D) pH of solution at half equivalence point.

At half equivalence point, half quantity of acid will be converted into its salt i e at this stage solution contain weak acid and it's salt, Hence this solution acts as a buffer and it's pH is calculated by using Henderson's equation.

pH = pKa + log [Salt] /[ Acid]

$\therefore$ pH = 4.87 + log  [CH₃CH₂COONa] /  [CH₃CH₂COOH]

We have, [CH₃CH₂COOH]= [CH₃CH₂COONa]

$\therefore$ pH = 4.87 + log 1

pH = 4.87 + 0

pH = 4.87

E)

scole x-axis, 16mm = 0.5 m Y-axis, community + 8 Volume of NaOH added