Determine the pH of an HF solution of each of the following concentrations. In which cases can you not make the simplifying assumption that x is small? (Ka for HF is 6.8×10−4.) a) 0.220 MExpress your answer to two decimal places.b) 0.0470 MExpress your answer to two decimal places.c) 0.0270 MExpress your answer to two decimal places.

Question
Asked Oct 23, 2019

Determine the pH of an HF solution of each of the following concentrations. In which cases can you not make the simplifying assumption that x is small? (Ka for HF is 6.8×10−4.) 

a) 0.220 M

Express your answer to two decimal places.

b) 0.0470 M
Express your answer to two decimal places.
c) 0.0270 M
Express your answer to two decimal places.
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Expert Answer

Step 1

(a)

ICE table for the reaction is given below.

HF H
F
Initial
0.220
0
0
Change
Equilibrium 0.220 -x
-X
X
X
X
X
н|F
[HF
!=6.8x10
к,-
-=6.8x10
0.220 x
x2+6.8x10-1.496x10= 0
x= 0.013
pHlogH
--log[0.013
1.89, pH of the solution.
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Image Transcriptionclose

HF H F Initial 0.220 0 0 Change Equilibrium 0.220 -x -X X X X X н|F [HF !=6.8x10 к,- -=6.8x10 0.220 x x2+6.8x10-1.496x10= 0 x= 0.013 pHlogH --log[0.013 1.89, pH of the solution.

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Step 2

(b)

ICE table for the reactio...

HF
H
F
Initial:
0.0470
0
0
Change
-X
X
X
Equilibrium 0.0470-x
X
HF
к-
=6.8x10
HF
-= 6.8x10
0.0470 x
x26.8x10x-3.196x105= 0
x = 0.006
pHlog H
--log [0.006]
-2.22, pHof the solution.
help_outline

Image Transcriptionclose

HF H F Initial: 0.0470 0 0 Change -X X X Equilibrium 0.0470-x X HF к- =6.8x10 HF -= 6.8x10 0.0470 x x26.8x10x-3.196x105= 0 x = 0.006 pHlog H --log [0.006] -2.22, pHof the solution.

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