Question

For 200.0 mL of a buffer solution that is 0.2975 M in CH3CH2NH2 and 0.2730 M in CH3CH2NH3Cl, calculate the initial pH and the final pH after adding 0.0100 mol of NaOH(Kb=6.46⋅10−4).

Answer 1

Given :

This is buffer so we use Henderson equation

pOH = pkb + log ( [conjugate acid ]/[conjugate base])

pOH = -log kb + log ( [conjugate acid ]/[conjugate base])

pOH = -log ( 6.46 E-4) + log ( [0.2730 ]/[0.2975])

= 3.15

pH = 14- 3.15 = 10.85

pH after addition of 0.0100 mol NaOH

Calculation of moles of conjugate acid and base

Moles of conj acid = CH3CH2NH3+ = Volume (L) x molarity

= 0.200 L x 0.2730 = 0.0546 mol

moles of base = 0.200 L x 0.2975 M

=0.0595 mol

We know the addition of NaOH will increase 1 mol base and decrease one mole acid. We can show it by using reaction

NaOH + CH3CH2NH3+ --à H2O + CH3CH2NH2

Then final moles of acid

CH3CH2NH3+ = Intial moles – moles of NaOH

= 0.0546 mol – 0.0100 mol = 0.0446 mol acid

Moles of base

= Initial moles + moles of NaOH

= 0.0595 mol + 0.0100 mol

= 0.0695 mol base

Volume is same so we can use moles as in the log ratio

Lets use write Henderson equation

pH = - log ( 6.46E-4) + log ( 0.0446/ 0.0695 )

= 2.997

pOH = 14 -2.997 = 11.0

pH increased by addition of base

and pH = 11.0