IIT JEE 2009 P2 Q14

The dissociation constant of a substituded benzoic acid at 25 degree C is 10-4

The pH of a 0.001 M solution of its sodium salt is

10 Answers

62
Lokesh Verma ·

[url=http://iitjee2009solutions.targetiit.com]Solution Key with Question Paper of 2009 IIT JEE[/url]

1
Manas ·

i m getting 6 as answer

11
rkrish ·

@bhaiyya......."The pH of a 0.01 M solution of its sodium salt is " not 0.001

I got pH = 4

Ka = 10-4
c = 0.01 M

α = 0.01

[H+] = 10-4
pH = 4

1
Thedarkknight ·

Answer is 8

1
sayantan27 ·

i also got 4 as my answer

1
Nishant Nemani ·

Ph of salt of weak acid and strong base
pH=7+.5[pka+log c]

so in this case pKa=4 log c= log .01=-2

pH= 7 + .5{4-2}=7+1=8

1
souvik seal ·

friends u r to consider H+ coming from water also...an acid with pH 8/////////////
never happen.,.,.,.,.,.,.,.,., simple ionic concept,.,consult books

6
AKHIL ·

it wud be 6 point somethin....

6
Kalyan IIT-K Beware I'm coming ·

nishant bhaiya since when did you start asking chemistry questions??:P

62
Lokesh Verma ·

:P

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