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0.1 mol hcl is dissolved in distilled water of volm v then at lim v tend to infinity( ph)soln= ? 2- percentage ionisation of water at certain temp is 3.6 into (10) to the power -7 % find kw and ph of water 3- what concn of ac- ions will reduce h3o+ ion to 2* 10 to the power -4 m in o.4 m soln of hac ka (hac)=1.8* 10 to the power -5

0.1 mol hcl is dissolved in distilled water of volm v then at lim v tend to infinity( ph)soln= ?
2- percentage ionisation of water at certain temp is 3.6 into (10) to the power -7 % find kw and ph of water
3- what concn of ac- ions will reduce h3o+ ion to 2* 10 to the power -4 m in o.4 m soln of hac ka (hac)=1.8* 10 to the power -5

Grade:12

1 Answers

Aarti Gupta
askIITians Faculty 300 Points
10 years ago

Dear student kindly post your questions one by one.
Sol.1:- 0.1M HCl dissolved in V of H2O.
Since V --------> infinity,thus solution becomes almost neutral.
Thus,pH of neutral sol. is 7.
Sol.2:- 3.6* 10-7% of H2O means 3.6*10-7 m ionizes out of 100m of H2O.
Thus, H2o ---> H+ + OH-
Kw = [H+ ] [ OH- ]
= 3.6 *10-7 * 3.6*10-7
12.96 *10-14
Now for pH = - Log [H+]
= - log(3.6*10-7)
= 6.44
Sol.3:-
CH3COOH --> CH3COO- + H3O+
Ka= [H3O][CH3COO-]/[CH3COOH]
1.8*10-5 = (2*10-4)[CH3COO-] / (0.4)
[CH3COO-] =1.8*10-5 * 0.4 / 2*10-4
=0.036 m

Thanks & regard

Aarti Gupta

askiitians Faculty

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