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In: Chemistry

For 530.0 mL of a buffer solution that is 0.155 M in CH3CH2NH2 and 0.145 M...

For 530.0 mL of a buffer solution that is 0.155 M in CH3CH2NH2 and 0.145 M in CH3CH2NH3Cl, calculate the initial pH and the final pH after adding 0.020 mol of HCl.

Expert Solution

V = 530 ml

M = 0.155 CH3CH2NH2

M = 0.145 CH3CH2NH3Cl

a) find pH initial

this is abuffer

its modeled with henderson haselback equation

pOH = pKb + log(HB+ / B)

Kb = 5.0*10^-4

pKb = -log(Kb) =-log(5.0*10^-4)= 3.3

pOH = 3.3 + log(0.145/0.155 )

pOH = 3.27

but we need pH so

pH = 14-pOH = 14-3.27 = 10.73

pH =10.73

b) after 0.02 mol of HCl

We will react base (neutralize it)

CH3CH2NH2 + HCl ---> H2O + CH3CH2NH3Cl

ratio is 1:1 so 1 mol of acid reacts with 1 mol of base to form salt + h2o

The moles before reaction

mol of CH3CH2NH2 = M*V = 0.155*¨0.53 = 0.08215

mol of CH3CH2NH3Cl = M*V = 0.145*0.53 = 0.07685

mol of HCl = 0.02

After reaction:

mol of CH3CH2NH2 =0.08215 - 0.02 = 0.06215

mol of CH3CH2NH3Cl = 0.07685 + 0.02 = 0.09685

mol of HCl = 0.02 - 0.2 = 0

there is no need to recalculate concnetrations since it is the same volume i.e. M = M and n = n since V = V

therefore

pOH = 3.3 + log(0.09685/0.06215 ) = 3.3 + 0.1926 = 3.49

pH =14 -pOH = 14-3.49 = 10.50

queen_honey_blossom answered 1 month ago

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