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The ionization constant of nitrous acid is 4.5xx10^(-4) . Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis . |
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Answer» Solution :CALCULATE pH of salt sodium nitrite solution :Nitrous Acid and sodium Nitrite combine and from sodium nitrite. This mixture is acidic buffer solution. DEGREE of hydrolysis (h) of `NaNO_2` (Sodium nitrite) : Ionic equilibrium in nitrous acid and `K_a` `HNO_(2(AQ)) hArr H_((aq))^(+) + NO_(2(aq))^(-)` `K_a=([H^+][NO_2^-])/([HNO_2])=4.5xx10^(-4)` Hydrolysis and CONSTANT of hydrolysis of `NO_2^(-)` : `{:(NO_2^(-) + H_2O hArr , HNO_2+ ,OH^-),((C-x)=C, x=C , x=C),((0.04),,):}` `K_h=([HNO_2][OH^-])/([NO_2^-])` `=((x)(x))/0.04 =x^2/0.04` ...(a) Put `(H^+)` in Numerator and Denominator in this expression , So, `K_h=([HNO_2][OH^-][H^+])/([NO_2^-][H^+])` But `[OH^-][H^+]=K_w` and `([HNO_2])/([NO_2^-][H^+])=1/K_a` `therefore K_h=K_w/K_a` `=(1.0xx10^(-14))/(4.5xx10^(-4))=2.222xx10^(-11)` Put value of `K_h` is equation (a), `2.222xx10^(-11)=x^2/0.04` `therefore x=sqrt(0.04xx2.222xx10^(-11))` `=sqrt(0.8888xx10^(-12))` `=0.9428 xx10^(-6)` `=9.428xx10^(-7)` =C `therefore h=x/C` `=(9.428xx10^(-7))/0.04` `=235.7xx10^(-7)` `=2.357xx10^(-5)` |
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