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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 water3- 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

```
6 years ago

## Answers : (1)

Aarti Gupta
300 Points
```

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*10-712.96 *10-14Now for pH = - Log [H+]= - log(3.6*10-7)= 6.44Sol.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

```
6 years ago
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