Question

For each of the following solutions, calculate the initial pH and the final pH after adding 0.005 mol of NaOH.

A.280.0 mL of pure water

initial pH, final pH

B.280.0 mL of a buffer solution that is 0.245 M in HCHO2 and 0.280 M in KCHO2

initial pH, final pH

C.280.0 mL of a buffer solution that is 0.320 M in CH3CH2NH2 and 0.300 M in CH3CH2NH3Cl

initial pH, final pH

Answer 1

A]

Initital pH = 7 ---> pure water

After adding 0.005 mol of NaOH

[OH-] = Moles / V in L = 0.005 / 280*10^-3 = 0.0178

pOH = -log[OH-] = - log 0.0178 = 1.75

pH = 14 - pOH = 12.25

B]

Initial pH :

pH = pKa + log [HCOOK] / [HCOOH]

pH = 3.75 + log [0.28 / 0.245] = 3.8

After adding NaOH

Moles of HCOOH = Molarity*V in L = 280*0.245/1000 = 0.0686

MOles of HCOOK = Molarity*V in L = 0.28*280/1000 = 0.0784

HCOOH + OH- ------------> HCOO-

0.0686 0.005 0.0784

0.0686-0.005 0 0.0784+0.005

pH = pKa + log [HCOO- ] / [HCOOH]

Substituting

pH = 3.867

C]

Initial pH

Buffer formula ---> basic buffer

pOH = pKb + log [Salt] / [Base]

pOH = 3.19 + log [ 0.3] / [0.32] { pKb of CH3CH2NH2 = 3.19 }

pOH = 3.162

pH = 14-pOH = 10.838

After adding NaOH

Moles of CH3CH2NH2 = 280*0.32 / 1000 = 0.0896

MOles of CH3CH2NH3Cl = 0.084

CH3CH2NH2 --------> CH3CH2NH3Cl + NaOH

0.0896 0.084 0.005

0.0896+0.005 0.084-0.005 0

pOH = pKb + log [0.084-0.005] / [0.0896+0.005]

pOH = 3.11

pH = 10.89