Question

Calculate the pH of each of the solutions and the change in pH to 0.01 pH units caused by adding 10.0 mL of 3.01-M HCl to 630. mL of each of the following solutions.

Change is defined as final minus initial, so if the pH drops upon mixing the change is negative.

a) water

pH before mixing =
pH after mixing=
pH change =

b) 0.109 M C₂H₃O₂^1-

pH before mixing =
pH after mixing=
pH change =

c) 0.109 M HC₂H₃O₂

pH before mixing =
pH after mixing=
pH change =

d) a buffer solution that is 0.109 M in each C₂H₃O₂^1- and HC₂H₃O₂

pH before mixing =
pH after mixing=
pH change =

Answer 1

Solution:- (a) Originally we have 630. ml water. Pure water has equal concentrations of [H⁺] and [OH^-], So,

the pH before mixing = 7

10.0 ml of 3.01 M HCl is added to 630. ml of water. It will decrease the concentration of HCl solution.

We can calculate it by using the equation, M₁V₁ = M₂V₂

V₁ and V₂ are initial and final volume. M₁ and M₂ are initial and final molarities. V₂ = 10.0 ml + 630. ml = 640. ml

3.01 M x 10.0 ml = M₂ x 640.ml

M₂ = 3.01 M x 10.0 ml/640.ml

M₂ = 0.0470M

This M₂ is also the concentration of [H⁺] since HCl a strong acid, gives H⁺ ion to the solution.

pH = - log(0.0470)

pH = 1.33 (this is the pH after mixing)

pH change = 1.33 - 7.00

pH change = -5.67

(b) Originally we have 0.109 M CH₃COO^-(acetate ion). It is a conjugate base of weak acid(acetic acid). We will write its hydrolysis equation and do the ice table to find out the pH as..

CH₃COO^-(aq) + H₂O(l) <--->   CH₃COOH(aq) + OH^-(aq)

I 0.109 0 0

C -X +X +X

E (0.109 - X) X X

Kb = [(X)²]/(0.109 - X)

Ka for acetic acid is 1.8 x 10^-5. So, kb = 1.0 x 10^-14/1.8 x 10^-5 = 5.56 x 10^-10

Kb value is very low so we could use the approxymation and the X on the bottom could be neglected.

5.56 x 10^-10 = [(X)²]/(0.109)

X = square root of (5.56 x 10^-10 x 0.109)

X = 7.78 x 10^-6

This X is also the value of OH^- (from ice table)

pOH = - log(7.78 x 10^-6)

pOH = 5.11

pH = 14 - 5.11

pH = 8.89 (pH before mixing)

moles of HCl added = 10.0 ml x (1L/1000ml) x (3.01mol/L) = 0.0301 mol

initial moles of CH₃COO^- = 630. ml x (1L/1000ml) x (0.109 mol/L) = 0.06867 mol

acetate ion will react with H⁺(from HCl) to form acetic acid and the solution will act as a buffer.

HCl is limited so the moles of CH₃COOH formed will be = 0.0301

excess moles of CH₃COO^- = 0.06867 - 0.0301 = 0.03857

pH is calcuated using handerson equation, pH = Pka + log(A^-/HA)

where HA is the weak acid and A^- is its conjugate base.

Pka for acetic acid is 4.74. So,

pH = 4.74 + log(0.03857/0.0301)

pH = 4.74 + 0.11

pH = 4.85(after mixing)

pH change = 4.85 - 8.89

pH change = -4.04

(c) pH of 0.109 M CH₃COOH (acetic acid) is also calculated using ice table since it is a weak acid.

CH₃COOH(aq) + H₂O(l) <-----> CH₃COO^-(aq) + H₃O⁺(aq)

I 0.109 0 0

C -X +X +X

E (0.109 - X) X X

1.8 x 10^-5 = [(X)²]/(0.109 - X)

value value is very low so the approxymation could also be used here.

1.8 x 10^-5  = [(X)²]/(0.109)

X = square root of (1.8 x 10^-5 x 0.109)

X = 1.40 x 10^-3

pH = - log(1.40 x 10^-3)

pH = 2.85 (pH before mixing)

intial moles of acetic acid are 0.06867 and 0.301 moles of HCl. Total volume will be 640. ml that is 0.640 L.

so, concentration of acetic acid in solution = 0.06867 mol/0.640L = 0.107 M

concentration of HCl or H⁺ = 0.0301 mol/0.640 L = 0.0470 M

we will make the ice table. Equilibrium will shift to the left side according to the Le Chatelier's principle...

CH₃COOH(aq) <----------> CH₃COO^-(aq) + H⁺(aq)

I 0.107 0 0.047

E +X -X -X

C (0.107 + X) X (0.047 - X)

1.8 x 10^-5  = (X)(0.047 - X)/(0.107 + X)

on solving this for X

X = 0.0000410

So, H⁺ = 0.047 - 0.0000410 = 0.046959

pH = - log(0.046959)

pH = 1.33 (after mixing)

change in pH = 1.33 - 2.85

change in pH = -1.52

(d)

Initially we have a buffer solution so the pH could be calculated using Handerson equation..

pH = Pka + log(A^-/HA)

pH = 4.74 + log(0.109/0.109)

pH = 4.74 + 0

pH = 4.74(before mixing)

Initial moles of acetic acid = 0.06867

initial moles of acetate ion = 0.06867

moles of HCl added = 0.0301

added HCl will react with acetate ion and form acetic acid. since HCl is limiting. So, moles of acetate ion aftre mixing will be = 0.06867 - 0.0301 = 0.03857

moles of acetic acid afrer adding HCl = 0.06867 + 0.0301 = 0.09877

new concentration of acetic acid = 0.09877 mol/0.640L = 0.154 M

new concentration of acetate ion = 0.03857 mol/0.640L = 0.0603 M

pH = 4.74 + log(0.0603/0.154)

pH = 4.74 - 0.41

pH = 4.33 (aftre mixing)   

pH change = 4.33 - 4.74

pH change = -0.41