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The ionization constant of benzoic acid is 6.46xx10^(-5) and K_(sp) for silver benzoate is 2.5xx10^(-13) . How many times is silver benzoate more soluble in a buffer of pH 3.19 comparedto its solubility in pure water ? |
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Answer» Solution :Solubility of silver benzoate in pure water (S) : `{:(C_6H_5COOAg_((s)) hArr , C_6H_5COO_((aq))^(-) + , Ag_((aq))^(+)),(S,"S M","S M"):}` `K_(sp)=[C_6H_5COO^-][Ag^+]=(S)(S)=S^2` `therefore S=SQRT(K_(sp))=sqrt(2.5xx10^(-13))=sqrt(25xx10^(-14))` `=5.0xx10^(-7)` `C_6H_5COOAg`, the solubility in pure water `[H^+]` in BUFFER (pH=3.19): pH=-log `[H^+]`=3.19 so log `[H^+]` = -3.19 `therefore [H^+]`= Antilog (-3.19)= `6.4566xx10^(-4)` M `approx 6.46xx10^(-4)` M Ionicequilibrium and CONCENTRATION in Benzoic acid : `C_6H_5COOH_((aq)) + H_2O hArr C_6H_5COO_((aq))^(-) + underset(6.46xx10^(-4))(H_3O_((aq))^(+))` `K_a=([C_6H_5COO^-][H_3O^+])/(C_6H_5COOH)` `therefore ([C_6H_5COOH])/([C_6H_5COO^-])=([H_3O^+])/K_a=(6.46xx10^(-4))/(6.46xx10^(-5))=10.0` `therefore [C_6H_5COOH]=10(C_6H_5COO^-)` Solubility of silver benzoic in Buffer (x) : Suppose solubility of silver benzoate in buffer (Benzoic acd + Silver benzoate) = x mol `L^(-1)` `(i)underset"(x)"(C_6H_5COOHAg_((s)))hArr, underset"x M"(C_6H_5COO_((aq))^(-))+underset"x M"(Ag_((aq))^(+))` (ii)`C_6H_5COOH+H_2O hArr C_6H_5COO_((aq))^(-) + H_3O_((aq))^(+)` Equilibrium (ii), produce `C_6H_5COOH` by the effect of common ion `C_6H_5COO^-` and REACTION is reverse . `therefore [Ag^+]=x=[C_6H_5COO^-]=[C_6H_5COOH]` but According to (i)`[C_6H_5COOH]=10[C_6H_5COO^-]` So, `x=[C_6H_5COO^-]+10[C_6H_5COO^-]` `=[C_6H_5COO^-](1+10)` `[Ag^+]=x=11[C_6H_5COO^-]` `therefore` in Buffer `[C_6H_5COO^-]=x/11` Calculation of x : `C_6H_5COOAg_((s)) hArr C_6H_5COO^(-) + Ag^+` here `[Ag^+] =x` and `[C_6H_5COO^-]=x/11` `therefore K_(sp)=[C_6H_5COO^-][Ag^+]` `therefore 2.5xx10^(-13) =(x/11) (x) =x^2/11` `therefore x = sqrt((2.5xx10^(-13))xx11)` `=sqrt(2.5xx1.1xx10^(-12))` `=sqrt(2.75xx10^(-12))` `=1.6583xx10^(-6)` `approx 1.66xx10^(-6)` M solubility in buffer Solubility in water =S= `5.0xx10^(-7)` = less VALUE Solubility in buffer = x= `1.66xx10^(-6)` M = more value `therefore "Solubility of silver benzoate in buffer"/"Solubility of silver benzoate in water"=(1.66xx10^(-6))/(5.0xx10^(-7))` =3.32 Thus, 3.32 times more solubility of silver benzoate in buffer . |
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