a)write a chemical equation for the reaction of this acid with water. Identify t
ID: 827871 • Letter: A
Question
a)write a chemical equation for the reaction of this acid with water. Identify the Bronsted acids and bases in the reaction.
b) Determine the Ka of this acid
c) if 250.0 ml of this weak acid is titrated with 1.00 M NaOH, what is the pH after 25.0 ml of the base is added?
d) what is the pH of the solution at the equivalence point of the titration described in C above?
e) A buffer solution is created using 0.15 moles of a salt NaB ('B' being the anion of the weak acid HB) and 50.0 ml of 2.0M acid HB ( assume no change in volume) . what would be the pH of that solution?
Explanation / Answer
The reaction is given by
HB + H20 -----> H30+ + B-
here the bronsted base is H20 and bronsted acid is HB
for a weak acid
H+ = sqrt ( KaC)
given pH = 3.2
[H+] = 6.309 x 10-4
6.309 x 10-4 = sqrt ( Ka x 0.50)
Ka = 7.96 x 10-7
so the value of Ka is 7.96 x 10-7
C) the reaction is given by
HB + NaOH = NaB + H20
moles of HB = 0.5 x 250/1000 = 0.125
moles of NaOH = 1 x 25 /1000 = 0.025
moles of NaB = moles of NaOH reacted = 0.025
moles of HB left = 0.125-0.025 = 0.10
for acidc buffer
pH = pKa + log (salt /acid )
so pH = pKa + log [ NaB / HB]
pH = -log 7.96 x 10-7 + log ( 0.025 /0.1)
pH = 5.50
so the pH is 5.50
d) at equivalence point
moles of acid = moles of base = 0.125
volume of base = 0.125 x 1000/1 = 125 ml
moles of NaB = 0.125
total volume = 250 + 125 = 375
conc of NaB = 0.125 x 1000/375 = 0.33
B- + H20 -----> OH- + HB
[OH-] = sqrt ( Kb C)
Kb = Kw /ka = 10-14 / 7.96 x 10-7 = 1.256 x 10-8
[OH-] = sqrt ( 1.256 x 10-8 x 0.33)
[OH-] = 6.47 x 10-5
pOH = -log 6.47 x 10-5
pOH = 4.19
pH = 9.81
so the pH is 9.81
e) moles of HB = 50 x 2 /1000 = 0.1
moles of NaB = 0.15
for acidic buffer
pH = pKa + log [NaB/HB]
pH = -log 7.96 x 10-7 + log (0.15/0.1 )
PH = 6.275
so the pH is 6.275
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