Ionic Equilibrium-1

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What is the pH of a solution prepared by mixing 0.0100 mol HA (with Ka = 1.00*10^-2) and 0.0100 mol of A^- and diluting with water to 1.00 L?

#Chemistry #Physical Chemistry

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6 Answers

1

sakshi pandey pandey ·2009-12-17 22:12:28

is the ans 2 sir???

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1

mentor_Layak Singh ·2009-12-17 22:23:13

How did you get ?
show ur calculations !!!

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1

sakshi pandey pandey ·2009-12-17 23:16:26

i used the formula
pH = pKa + log[ A-]/ [ HA]

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mentor_Layak Singh ·2009-12-18 00:05:28

no, you can't direct put up the values in formula.

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106

Asish Mahapatra ·2009-12-18 00:13:19

HA = H+ + A-
0.01 0.01

t=teq. 0.01(1-x) 0.01x 0.01(1+x)

Ka = [H+][A-]/[HA]

=> 10^-2 = 0.01x*0.01(1+x)/0.01(1-x)

=> 10^-2 = 10^-2 (x+x²)/(1-x)

=> 1-x = x+x²
=> x²+2x-1=0

=> x = âˆš2 -1

So pH = -log(0.01x)
= 2 - log(x)
= 2 - log(0.414)
= 2-(-0.38)
= 2.38

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mentor_Layak Singh ·2009-12-18 01:16:21

Yes !!
Your answer is correct !!!

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