In: Chemistry
A 25.0 mL sample of 0.125 molL−1 pyridine (Kb=1.7×10−9) is titrated with 0.100 molL−1HCl.
Part A
Calculate the pH at 0 mL of added acid.
Express your answer using two decimal places.
Part B
Calculate the pH at 10 mL of added acid.
Express your answer using two decimal places.
Part C
Calculate the pH at 20 mL of added acid.
Express your answer using two decimal places.
Part D
Calculate the pH at equivalence point.
Express your answer using two decimal places.
Part E
Calculate the pH at one-half equivalence point.
Express your answer using two decimal places.
Part F
Calculate the pH at 40 mL of added acid.
Express your answer using two decimal places.
Part G
Calculate the pH at 50 mL of added acid.
Express your answer using two decimal places.
A 25.0 mL sample of 0.125 molL−1 pyridine (Kb=1.7×10−9) is titrated with 0.100 molL−1HCl.
a)
initially
Kb = [HB+][OH-]/[B]
1.7*10^-9 = x*x/(0.125-x)
x = 1.457*10^-5
[OH-] = 1.457*10^-5
pOH = -log(1.457*10^-5) = 4.8365
pH = 14-4.8365 = 9.1635
b)
after 10 ml of acid
mmol of H+ = MV = 10*0.1 = 1 mmol
mmol of base = 0.125*25 = 3.125 mmol
mmol of base left = 3.125-1 = 2.125
mmol of conjugate = 0+ 1
this is a buffer
so
pOH = pKb + log(HB+/B)
pOH = -log(1.7*10^-9) + log(1/2.125) = 8.4421
pH = 14-8.4421 = 5.5579
c)
V = 20 mL
mmol of H+ = MV = 20*0.1 = 2 mmol
mmol of base = 0.125*25 = 3.125 mmol
mmol of base left = 3.125-2 = 1.125
mmol of conjugate = 0+ 2
this is a buffer
so
pOH = pKb + log(HB+/B)
pOH = -log(1.7*10^-9) + log(2/1.125) = 9.01942
pH = 14-9.01942= 4.98058
d)
in equivalence point
mmol of base = MV = 25*0.125 = 3.125
volume of acid = mmol/M = 3.125/0.1 = 31.25 mL
totla V= 31.25+25 = 56.25 mL
so..
[HB+]= m1*v1 = 3.125 /56.25 = 0.05555
Ka = [H+][B]/[HB+]
KA = Kw/Ka = (10^-14)/(1.7*10^-9) = 5.882*10^-6
5.882*10^-6 = x*x/(0.05555-x)
x = 5.65*10^-4
pH = -log( 5.65*10^-4
pH= 3.2479
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