Question
Asked Nov 21, 2019
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Calculate the pH of a solution formed by mixing 250.0 mL of 0.15 M NH4Cl with 100.0 mL of 0.20 M NH3. The Kb for NH3 is 1.8 x 10-5.

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Expert Answer

Step 1

Given information:

Volume of NH4Cl solution (V1) =250.0 mL

Volume of NH3 solution (V2) =100.0 mL

Concentration of NH4Cl solution (M1) = 0.15 M

Concentration of NH3 solution (M2) = 0.20 M

Step 2

First the concentration of new solution formed  is calculated as follows:

 

M, V (Salt) M,V (NH,C)
M2V3 (NH,CI)
М,
V (Salt)
0.15 Mx250 mL
(250+100) mL
0.15 Mx250 mL
300 mL
= 0.125 M
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M, V (Salt) M,V (NH,C) M2V3 (NH,CI) М, V (Salt) 0.15 Mx250 mL (250+100) mL 0.15 Mx250 mL 300 mL = 0.125 M

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Step 3

Now, the concentration of base is calc...

M, V (Base) M2 V, (NH,)
M2V2 (NH3)
М,
V (Base)
0.20 M100 mL
(250+100) mL
0.20 M 100 mL
300 mL
=0.0666 M
help_outline

Image Transcriptionclose

M, V (Base) M2 V, (NH,) M2V2 (NH3) М, V (Base) 0.20 M100 mL (250+100) mL 0.20 M 100 mL 300 mL =0.0666 M

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