InterviewSolution
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InterviewSolution
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`500mL` of `0.2M` aqueous solution of acetic acid is mixed with `500mL` of `0.2HCI` at `25^(@)C`. a. Calculate the degree of dissociation of acetic acid in the resulting solution and `pH` of the folution. b. If `6g` of `NaOH` is added to the above solution determine the final `pH.[K_(a)` of `CH_(3)COOH =2 xx 10^(-5)`. |
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Answer» (a) Meq. Of `CH_(3)COOH=500xx0.2=100` Meq. Of `HCI=500xx0.2=100` `:. [HCI]=(100)/(1000)=0.1, [CH_(3)COOH]=(100)/(1000)=0.1` For `CH_(3)COOH:` `{:(,CH_(3)COOHhArr,CH_(3)COO^(-)+,H^(+)),("Before dissociation",0.1,0,0.1("from HCI")),("After dissociation", (0.1x),x,(0.1+x)):}` Due to common ion effect, dissociation of `CH_(3)COOH` is very small in presence of HCI. Therefore `(0.1+x)=0.1 and (0.1-x)=0.1`. `:. K_(a)=(x xx0.1)/(0.1)` `:. x=K_(a)=1.75xx10^(-5)` Thus degree of dissociation `alpha= (x)/(0.1)=(1.757xx10^(-5))/(0.1)` `=1.75xx10^(-4)` `=0.000175=0.0175%` Also `[H^(+)]=0.1+x=0.1` , `(.: xltltalpha1)` `:. pH= -log[H^(+)]= -log[0.1]=1` (b) Meq. of NaOH mole of NaOH added `= (6)/(40)=0.15` Therefore new equilibrium will have `{:(CH_(3)COOH+,HCI+,NaOHrarr,CH_(3)COONa+,NaCI+,H_(2)O),(0.1,0.1,0.15,0,0,0),(0.5,0,0,0.05,0,0):}` Thus the solution will act as buffer having `[CH_(3)COOH]` `=(0.5)/(1000)` and `[CH_(3)COONa]=(0.5)/(1000)` Thus, `pH= -log K_(a)+log ((["Salt"])/(["Acid"]))` `= -log 1.75xx10^(-5)+log (([0.05//1000])/([0.05//1000]))` `pH=4.757` |
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