Question
Asked Oct 16, 2019

A buffer solution contains 0.462 M CH3NH3Br and 0.381 M CH3NH2 (methylamine). Determine the pH change when 0.082 mol KOH is added to 1.00 L of the buffer.

pH after addition − pH before addition = pH change = ______

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Expert Answer

Step 1

pKb of CH3NH2 is 3.36.

pOH of the solution can be determined using Henderson-Hasselbalch equation. pH can be determined from the pOH value as follows,

Henderson Hasselbalch expression:
[Conjugate acid]
Base
РОН 3 рК, +1оg
[CH,NH,Br
[CH,NH2
(0.462 M)
' (0.381 M
=3.36 log
= 3.36+log
= 3.44
PH + pОН %3D 14
pH 3 14-рОН
14-3.44
10.56
Hence, pH of the solution before adding KOH is 10.56
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Henderson Hasselbalch expression: [Conjugate acid] Base РОН 3 рК, +1оg [CH,NH,Br [CH,NH2 (0.462 M) ' (0.381 M =3.36 log = 3.36+log = 3.44 PH + pОН %3D 14 pH 3 14-рОН 14-3.44 10.56 Hence, pH of the solution before adding KOH is 10.56

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Step 2

On addition of KOH, it reacts with CH3NH3Br forming CH3NH2. Hence, concentration of CH3NH3Br decreases whereas concentration of CH3NH2 increases after the addition of KOH. Hence, the concentration of CH3NH3Br and CH3NH2 is calculated.

CH,NH OH -
> CH,NH2H0
Concentration of KOH No. of moles x Volume(in L)
(0.082 mol)(1.00 L) = 0.082 M
=
[CH,NH,Br after adding KOH= (0.462 M)-(0.082 M) 0.380 M
CHNH
after adding KOH = (0.381 M) +(0.082 M) = 0.463 M
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CH,NH OH - > CH,NH2H0 Concentration of KOH No. of moles x Volume(in L) (0.082 mol)(1.00 L) = 0.082 M = [CH,NH,Br after adding KOH= (0.462 M)-(0.082 M) 0.380 M CHNH after adding KOH = (0.381 M) +(0.082 M) = 0.463 M

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Step 3

pH of the solution after adding KOH ...

[Conjugate acid]
Base
РОН - рК, +1оg
CH,NH,Br]
[CH,NH2
(0.380 M)
=3.36+log
= 3.36 +log
= 3.27
'(0.463 M)
PH + pОН 3 14
рH — 14- рОН
= 14-3.27 = 10.73
Hence, pH of the solution after adding KOH is 10.73
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[Conjugate acid] Base РОН - рК, +1оg CH,NH,Br] [CH,NH2 (0.380 M) =3.36+log = 3.36 +log = 3.27 '(0.463 M) PH + pОН 3 14 рH — 14- рОН = 14-3.27 = 10.73 Hence, pH of the solution after adding KOH is 10.73

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