Question
Asked Oct 11, 2019
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Calculate the pH of a buffer obtained by mixing 125 mL of 0.10 M NH3 with 250 mL of 0.12 M NH4Cl. Kb of NH3=1.8 x 10-5.

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Expert Answer

Step 1

pKb of the buffer solution has to be determined for calculating the pOH of the solution.

pK log K
-log (1.8 x10
= 4.74
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pK log K -log (1.8 x10 = 4.74

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Step 2

Concentration of 0.10 M NH3 and 0.12 M NH4Cl in the solution is determined using the formula form molarity. Total volume of the solution after mixing is 125 mL + 250 mL = 375 mL = 0.375 L.

1 mL = 0.001 L

No. of moles
Molarity
Volume
Before mixing
No. of moles of NH, (0.10 M)(0.125 L)= 0.0125 mol
No. of moles of NH Cl (0.12 M) (0.250 L) 0.03 mol
After mixing:
0.0125 mol
Concentration of NH
=0.033 M
0.375 L
0.03 mol
Concentration of NH,Cl=
= 0.08 M
0.375 L
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No. of moles Molarity Volume Before mixing No. of moles of NH, (0.10 M)(0.125 L)= 0.0125 mol No. of moles of NH Cl (0.12 M) (0.250 L) 0.03 mol After mixing: 0.0125 mol Concentration of NH =0.033 M 0.375 L 0.03 mol Concentration of NH,Cl= = 0.08 M 0.375 L

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Step 3

pOH of the solution can be determined usi...

Henderson Hasselbalch expression
Conjugate acid
Base
pOH pKog
[NH,CI]
NH,
= 4.74 log
(0.08 M)
(0.033 M)
= 4.74 log
= 5.12
Hence, pOH 5.12 for the buffer solution
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Henderson Hasselbalch expression Conjugate acid Base pOH pKog [NH,CI] NH, = 4.74 log (0.08 M) (0.033 M) = 4.74 log = 5.12 Hence, pOH 5.12 for the buffer solution

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