Weak diprotic acid H2A is titrated with 0.2000 M NaOH. The initial concentration

Question

Weak diprotic acid H2A is titrated with 0.2000 M NaOH. The initial concentration of H2A is 0.1000 M and the initial volume is 50.00 mL. Ka1 = 1.0 x 10?5 Ka2 = 1.0 x 10?9 a) Identify the predominate species at the first equivalence point, by chemical formula: b) Calculate the pH of the solution before any base is added. c) Calculate the pH at the first equivalence point. d) How many mL base, total, must be added to achieve the 2nd equivalence point? e) Calculate the pH at the second equivalence point. f) ) From the attached table of acid/base indicator transition ranges (bottom half of p.7), identify an appropriate indicator for: i) the 1st equivalence point ii) the 2nd equivalence point:

Explanation / Answer

a) H2A + OH-(aq) <--------> HA-(aq) + H2O  

at first equivalence point dominant species is HA-(aq) i.e NaHA

b) before base addition

H2A <-----> HA- + H+ , at equi [H2A] = ( 0.1x0.05-x)/0.05 , [H+]=[HA-] = x/0.05

Ka =[HA-][H+]/[H2A] , 10^-5 = (x/0.05)^2/(0.005-x)/0.05

x= 0.00004975 = HA- moles , [HA-] =[H+] = ( 0.00004975/0.5) = 0.0000995

now HA- <------> H+ + A- , at equi [HA] = ( 0.00004975-x)/0.05 , [H+]=[A-] =x/0.05,

Ka = 10^-9 = (x/0.05)^2/( 0.00004975-x)/0.05,

x = H+ moles = 5 x10^-8 , [H+] = 5 x10^-8/0.05 = 10^-6

so total [H+] = ( 0.00004975+5x10^-8)/0.05 = 0.000996 , pH = -log(0.000996) = 3

c) at first equi piont NaOH moles = H2A moles = 0.05 , NaOH vol = 0.005/0.2 = 0.025 L,

H2A + OH-(from NaOH) <----> HA-(aq) + H2O

now back reaction favours at first equi point

HA- + H2O <----> H2A + OH-

equi [HA-] = ( 0.005-x)/(0.05+0.025) , [H2A]=[OH-] = x/0.075

Kb = Kw/Ka = 10^-14/10^-5 = 10^-9 = [H2A][OH-]/[H2A]

10^-9 = ( x/0.075)^2/( 0.005-x) /0.075

x = 6.124 x10^-7 =[OH-] , pOH = -log ( 6.124x10^-7) = 6.213

pH = 14-6.123 = 7.9

d) vol at 2n equivalence = 2 times first equivalence base vol = 2 x 0.025 = 0.05 liters = 50ml,

e) H2A + 2OH- <-----> A2- (aq) + 2H2O

A2-(aq) + H2O <--> HA-(aq) + OH- occurs ,

A2- moles = 2 x NaOH moles = 2 x ( 0.2x0.005) = 0.01 , at equi [A2-]= ( 0.01-x)/0.1 ,

[OH-]=[HA-]=x/0.1

Kb = 10^-14/10^-9 = 10^-5 = ( x/0.1)^2/(0.01-x)0.1

x = OH- moles = 0.0000995, [OH-] = ( 0.0000995/0.1) = 0.000995 , pOH = -log(0.00095) = 3

pH = 14-3 = 11

f) indicator for 1st equipoint is phenopthelin ( pH range 6-8)

ii) pH indicator for 2nd equi piont is alizarin yellow( pH range 10-12)

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