1) What is the hydroxide ion concentration, [OH?], at the equivalence point when

Question

1) What is the hydroxide ion concentration, [OH?], at the equivalence point when 20.0 mL of 0.100 M HNO2 is titrated with 0.100 M NaOH? The Ka for HNO2 is 4.5

Explanation / Answer

1) HNO2 + NaOH --> NaNO2 + H2O , at equivalnce point we have only NaNO2 , NO2- + H2O ---> OH- + HNO2 , thus we get OH_ ions since NaNO2 is salt of strong base and weak acid , Kb = Kw/Ka = 10^ -14/4.5 x10^ -4 = 2.2 x10^ -11, moles of NaNO2 = moles of acid = moles of base = ( 20x0.1/1000) = 0.002, [NO2-] = 0.002 x1000/(20+20) = 0.05 , at equi [NO2-] = 0.05-x , [OH-] = x =[HNO2] , Kb = [OH-][HNO2]/[NO2-] , 2.2 x10^ -11 = (x)^2/(0.05-x) , x = 1.049 x10^ -6 =[OH-] , 2) pH = pKa + log [sodium acetate]/[acetic acid] , pka = -log(1.8 x10^ -5) = 4.75 , now 4 = 4.75 + log[sodium acetate]/[acetic acid] , [sodium acetate]/[acetic acid] = 0.18

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