Saved Bookmarks
| 1. |
The ionisation constant of chloroacetic acid is 1.35xx10^(-3) .What will be the pH of 0.1 M acid and its 0.1 M sodium salt solution ? |
|
Answer» Solution :PH of 0.1 M chloroacetic ACID `(ClCH_2COOH)` : This is a weak acid and suppose x M of 0.1 M is ionised and equilibrium is established. So At equilibrium `x=[H_3O^+] = [ClCH_2COO^-]` `{:(,ClCH_2COOH_((aq)) +H_2O_((l)) hArr , ClCH_2COO_((aq))^(-) + , H_3O_((aq))^(+)),("At equilibrium" , 0.1 M, 0.0 , 0.0),(,(0.1 -x)M,x M, xM),(,approx 0.1, ,):}` `K_a=([ClCH_2COO^(-)][H_3O^+])/([ClCH_2COOH])` `therefore 1.35 xx 10^(-3) =((x)(x))/0.1` `therefore x^2=1.35xx10^(-4)` `therefore x=1.162xx10^(-2)M=[H_3O^+]` pH=log `[H^+]` `=-log (1.162xx10^(-2))` =-(0.0652-2) =1.9348 `approx` 1.93 `therefore` pH of this weak base is 1.93 Acid `ClCH_2COOH` and SALT `ClCH_2COONa` present in solution so solution is buffer. The pH of this buffer is calculate by following equation. `pH=pK_a + "log"([A^-])/([HA])` `[A^-]=[ClH_2COO^-]=0.1` [HA]=`[ClCH_2COOH]`=0.1 where `pK_a=-log (K_a)` =-log `(1.35xx10^(-3))` =2.8690 pH=2.869 +log `(0.1/0.1)`=2.869 |
|