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In: Chemistry

Calculate the PH after the addition of 0.00, 5.00, 15.00, 25.00, 40.00, 45.00,49.00, 50.00, 51.00, 55.00...

Calculate the PH after the addition of 0.00, 5.00, 15.00, 25.00, 40.00, 45.00,49.00, 50.00, 51.00, 55.00 and 60mL of 0.100M HCl in the tirtration of 50.00mL of 0.100M weak Monoprotic base B. Kb of B is 9.52 *10^-7

Solutions

Expert Solution

(a) No acid added

B + H2O <==> BH+ + OH-

let x amount has hydrolyzed

Kb = 9.52 x 10^-7 = x^2/0.1

x = [OH-] = 3.08 x 10^-4 M

pOH = -log[OH-] = 3.51

pH = 14 - pOH = 10.49

(b) after 5 ml of 0.1 M HCl

moles of base = molarity x volume = 0.1 x 0.05 = 5 x 10^-3 mols

moles of acid added = 0.1 x 0.005 = 5 x 10^-4 mols

So, 5 x 10^-4 mols of acid is neutralized by 5 x 10^-4 mols of base

remaining base = 5 x 10^-3 - 5 x 10^-4 = 4.5 x 10^-3 mols

molarity of salt = 5 x 10^-4/(0.005 + 0.05) = 9.1 x 10^-3 M

molarity of base = 4.5 x 10^-3/(0.005 + 0.05) = 0.082 M

Using Hendersen-Hasselbalck equation,

pH = pKa + log([base]/[acid])

Ka = Kw/Kb = 1 x 10^-14/9.52 x 10^-7 = 1.05 x 10^-8

pKa = -log[Ka] = 7.98

pH = 7.98 + log(0.082/9.1 x 10^-3)

      = 8.93

(c) after 15 ml of 0.1 M HCl

moles of base = molarity x volume = 0.1 x 0.05 = 5 x 10^-3 mols

moles of acid added = 0.1 x 0.015 = 1.5 x 10^-3 mols

So, 1.5 x 10^-3 mols of acid is neutralized by 1.5 x 10^-3 mols of base

remaining base = 5 x 10^-3 - 1.5 x 10^-3 = 3.5 x 10^-3 mols

molarity of salt = 1.5 x 10^-3/(0.015 + 0.05) = 0.023 M

molarity of base = 3.5 x 10^-3/(0.015 + 0.05) = 0.054 M

Using Hendersen-Hasselbalck equation,

pH = pKa + log([base]/[acid])

Ka = Kw/Kb = 1 x 10^-14/9.52 x 10^-7 = 1.05 x 10^-8

pKa = -log[Ka] = 7.98

pH = 7.98 + log(0.054/0.023)

      = 8.35

(d) after 25 ml of 0.1 M HCl

moles of base = molarity x volume = 0.1 x 0.05 = 5 x 10^-3 mols

moles of acid added = 0.1 x 0.025 = 2.5 x 10^-3 mols

So, 2.5 x 10^-3 mols of acid is neutralized by 2.5 x 10^-3 mols of base

This is half equivalence point

So, pH = pKa = 7.98

(e) after 40 ml of 0.1 M HCl

moles of base = molarity x volume = 0.1 x 0.05 = 5 x 10^-3 mols

moles of acid added = 0.1 x 0.040 = 4 x 10^-3 mols

So, 4 x 10^-3 mols of acid is neutralized by 4 x 10^-3 mols of base

remaining base = 5 x 10^-3 - 4 x 10^-3 = 1 x 10^-3 mols

molarity of salt = 4 x 10^-3/(0.04 + 0.05) = 0.044 M

molarity of base = 1 x 10^-3/(0.04 + 0.05) = 0.011 M

Using Hendersen-Hasselbalck equation,

pH = pKa + log([base]/[acid])

Ka = Kw/Kb = 1 x 10^-14/9.52 x 10^-7 = 1.05 x 10^-8

pKa = -log[Ka] = 7.98

pH = 7.98 + log(0.011/0.04)

      = 7.37

(f) after 45 ml of 0.1 M HCl

moles of base = molarity x volume = 0.1 x 0.05 = 5 x 10^-3 mols

moles of acid added = 0.1 x 0.045 = 4.5 x 10^-3 mols

So, 4.5 x 10^-3 mols of acid is neutralized by 4.5 x 10^-3 mols of base

remaining base = 5 x 10^-3 - 4.5 x 10^-3 = 5 x 10^-4 mols

molarity of salt = 4.5 x 10^-3/(0.045 + 0.05) = 0.047 M

molarity of base = 5 x 10^-4/(0.045 + 0.05) = 5.26 x 10^-3 M

Using Hendersen-Hasselbalck equation,

pH = pKa + log([base]/[acid])

Ka = Kw/Kb = 1 x 10^-14/9.52 x 10^-7 = 1.05 x 10^-8

pKa = -log[Ka] = 7.98

pH = 7.98 + log(5.26 x 10^-3/0.047)

      = 7.03

(g) after 49 ml of 0.1 M HCl

moles of base = molarity x volume = 0.1 x 0.05 = 5 x 10^-3 mols

moles of acid added = 0.1 x 0.049 = 4.9 x 10^-3 mols

So, 4.9 x 10^-3 mols of acid is neutralized by 4.9 x 10^-3 mols of base

remaining base = 5 x 10^-3 - 4.9 x 10^-3 = 1 x 10^-4 mols

molarity of salt = 4.9 x 10^-3/(0.049 + 0.05) = 0.0495 M

molarity of base = 1 x 10^-4/(0.049 + 0.05) = 1.01 x 10^-3 M

Using Hendersen-Hasselbalck equation,

pH = pKa + log([base]/[acid])

Ka = Kw/Kb = 1 x 10^-14/9.52 x 10^-7 = 1.05 x 10^-8

pKa = -log[Ka] = 7.98

pH = 7.98 + log(1.01 x 10^-3/0.0495)

      = 6.29

(h) after 50 ml of 0.1 M HCl

moles of base = molarity x volume = 0.1 x 0.05 = 5 x 10^-3 mols

moles of acid added = 0.1 x 0.05 = 5 x 10^-3 mols

This the equivalence point

molarity of salt = 5 x 10^-3/0.1 = 0.05 M

Salt hydrolyzes in water, let x amount has hydrolyzed then,

Kb = 9.52 x 10^-7 = x^2/0.05

x = [H+] = 2.18 x 10^-4

pH = -log[H+] = 3.66

(i) after 51 ml of 0.1 M acid is added

1 ml of acid is in excess

mols of acid = 0.1 x 0.001 = 1 x 10^-4 mols

molarity of acid = 1 x 10^-4/0.101 = 9.9 x 10^-4 M

pH = -log(9.9 x 10^-4) = 3.0

(j) after 55 ml of acid added

molarity of acid = (0.1 x 0.005)/(0.05 + 0.055) = 4.76 x 10^-3 M

pH = 2.32

(k) after 60 ml of acid added

molarity of acid = (0.1 x 0.010)/(0.05 + 0.060) = 9.1 x 10^-3 M

pH = 2.04

queen_honey_blossom answered 1 year ago
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