Question

# for each of the following Solutions calculate the initial pH in the final pH after adding...

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

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

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