Question

Calculate the pH change that results when 11 mL of 5.8 M NaOH is added to 776 mL of each the following solutions. (See the appendix.)

pKa = 9.25

Ka = 5.6e-10

(c) 0.10 M NH₃

Answer 1

given

Ka = 5.6 x 10-10

we know that

Ka x Kb = 10-14

so

5.6 x 10-10 x Kb = 10-14

Kb = 1.7857 x 10-5

now

NH3 is a weak base

for weak bases

[OH-] = sqrt (Kb x C)

given

conc of NH3 , C = 0.1

so

[OH-] = sqrt ( 1.7857 x 10-5 x 0.1 )

[OH-] = 1.336 x 10-3

we know that

pOH= -log [OH-]

pOH = -log 1.336 x 10-3

pOH = 2.874

we know that

pH = 14 - pOH

pH = 14 - 2.874

pH = 11.126

so the initial pH of NH3 solution is 11.126

now

11 ml of 5.8 M NaOH is added

we know that

moles = molarity x volume (L)

so

moles of NaOH added = 5.8 x 11 x 10-3

moles of NaOH added = 0.0638

NaOH ---> Na+ + OH-

from the above reaction

moles of OH- from NaOH = moles of NaOH = 0.0638

consider NH3

[OH-] = 1.336 x 10-3

moles of OH- from NH3 = 1.336 x 10-3 x 776 x 10-3

moles of OH- from NH3 = 1.036736 x 10-3

total moles of OH- = 0.0638 + (1.036736 x 10-3)

total moles of OH- = 0.0648367

final volume = 11 + 776 = 787 ml

we know that

molarity = moles / volume (L)

so

[OH-] = 0.0648367 / 787 x 10-3

[OH-] = 0.08238467

now

pOH = -log [OH-]

pOH = -log 0.08238467

pOH = 1.084

now

pH = 14 -pOH

pH = 14 - 1.084

pH = 12.916

so the final pH is 12.916

pH change = final - initial

pH change = 12.916 - 11.126

pH change = 1.79

so the change in pH is +1.79