Dissociated constant of weak acid `CH_3COOH` is `1.8xx10^(-5)` . In 0.1 M solution calculate concentration `CH_3COO^-` and `H^+` . Calculate pH of solution. If 0.1 M HCl added to this solution than calculate degree of dissociation of `CH_3COOH`.

Explanation

Calculation of `[H^+]` and `[CH_3COO^-]` : Suppose in initial 0.1 `CH_3COOH` is dissociated in `alpha` amount. So, dissociated `[CH_3COOH]=0.1 alpha` M and `[CH_3COO^-]=[H^+]=alphaM`
`{:("Equili.:",CH_3COOH_((aq))+H_2O_((l))hArr ,CH_3COO_((aq))^(-)+,H_3O_((aq))^(+)),("Initial:",0.1 M,0 M,0M),("Change in equilibrium :",-0.1 alpha,0.1 alpha,0.1 alpha),("Concen. of equili. :",=0.1(1-alpha),,):}`
`K_a=([CH_3COO^-][H_3O^+])/([CH_3COOH])`
`therefore 1.8xx10^(-5) ((alpha)(alpha)0.1^2)/(0.1(1-alpha))=(0.1alpha^2)/(1-alpha)`
But, `1 gt gt alpha` so, `(1-alpha)`=1 take
`1.8xx10^(-5)=0.1 alpha^2`
`therefore alpha^2 = 1.8xx10^(-4)`
`therefore alpha=1.3416xx10^(-2)`
`[H^+]=[CH_3COO^-]`
=`0.1alpha=0.1(1.3416xx10^(-2))`
`=1.3416xx10^(-3)`
Calculation of pH :
`pH=log [H^+]= -log (1.3416xx10^(-3))`
=-log (0.0013416)
=2.8724=2.87
Dissociation degree in 0.1 HCl :
`{:(HCl to, H_((aq))^(+) +, Cl_((aq))^(-)),(0.1 M,0.1 M,0.1M):}`
`{:(CH_3COOH_((aq)) (H_2O)hArr, H_3O_((aq))^(+)+, CH_3COO_((aq))^(-)),(,0.1 alpha,0.1 alpha) :}`
`therefore` Total `[H_3O^+]=[H_3O^+]` of HCl + `[H_3O^+]` of `CH_3COOH`
`=(0.1+0.1 alpha)`
`approx` 0.1 M (Because `alpha lt lt` 0.1)
Now , `K_a=([CH_3COO^-][H^+])/([CH_3COOH])`
`therefore 1.8xx10^(-5)=((0.1 alpha)(0.1))/(0.1(1-alpha))[ 1 gt gt gt alpha therefore 1-alpha approx 1]`
`1.8xx10^(-5)=0.1 alpha`
`therefore alpha = 1.8xx10^(-4)`