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

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^-
K_w = [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⁺
K_a= [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