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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. |
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Answer» SOLUTION :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)` |
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