Question

for each of the following Solutions calculate the initial pH in the final pH after adding 0.200 moles of NaOH for

300.0 M L of a buffer solution that is 0.2551 m in ch3ch2nh2 AND 0.2351 in ch3ch2nh3cl calculate the initial pH and the final pH after adding 0.020 moles of naoh KB equal 5.6 x 10 to the negative 4

Answer 1

Given that,

[CH3CH2NH2] = 0.2551

[CH3CH2NH3Cl] = 0.2351

Kb = 5.6*10^-4

pKb = - log Kb

= 3.25

We know that pKa+ pKb = 14

pKb = 14-3.25

= 10.75

Now use the Henderson Hasselbalch eqn. to calculate the pH

pH = pKa + log([CH3CH2NH2] / [CH3CH2NH3Cl])

= 10.75 + log(0.2551/0.2351)

= 10.75 + log 1.085

= 10.75 + 0.035

= 10.785

now calculate the Final pH after adding 0.0200 moles NaOH:

Number of mole s= volume in L * molarity

moles CH3CH2NH2 = 0.300 L x 0.2551 M = 0.07653
moles CH3CH2NH3Cl = 0.300 L x 0.2351M = 0.07053

CH3CH2NH3+ + OH- = CH3CH2NH2 + H2O

moles CH3CH2NH3+ = 0.07053 - 0.0200= 0.05053
moles CH3CH2NH2 = 0.07653+ 0.0200 = 0.09653
now calculate new molarity as follows:

[CH3CH2NH3+] = 0.05053/ 0.300 = 0.1684 M
[CH3CH2NH2] = 0.09653 /0.300 L = 0.3218 M
pH = 10.75 + log(0.3218 / 0.1684)

=10.75+ 0.281

= 11.031