w/SB Consider the titration of 40.0 ml of an aqueous solution of 0.250 M HF (a w
ID: 556838 • Letter: W
Question
w/SB Consider the titration of 40.0 ml of an aqueous solution of 0.250 M HF (a weak monoprotic acid: K,-3.5x 10-) with 0.200 M NaOH (a strong base). Calculate the pH of the solution at each of the following points: 17. Before any base has been added to the solution. 2.03 e) 3.14 a) 1.69 b) 2.78 c) 1.98 18. After the addition of 10.0 mL of base. d) 4.01 e) 2.97 a) 3.95 b) 3.44 c) 2.85 19. Halfway to the equivalence point (titration midpoint). a) 3.77 b) 4.23 c) 3.46 d) 4.89 e) 2.92 20. At the equivalence point. a) 8.25 d) 8.53 e) 7.45 b) 6.44 c) 7.00 21. After the addition of 80.0 mL of base. a) 11.42 b) 12.70 c) 10.95 d) 12.06 e) 13.18Explanation / Answer
Q17
before any base
First, assume the acid:
HF
to be HA, for simplicity, so it will ionize as follows:
HA <-> H+ + A-
where, H+ is the proton and A- the conjugate base, HA is molecular acid
Ka = [H+][A-]/[HA]; by definition
initially
[H+] = 0
[A-] = 0
[HA] = M;
the change
initially
[H+] = + x
[A-] = + x
[HA] = - x
in equilbrirum
[H+] = 0 + x
[A-] = 0 + x
[HA] = M - x
substitute in Ka
Ka = [H+][A-]/[HA]
Ka = x*x/(M-x)
x^2 + Kax - M*Ka = 0
if M = 0.25 M; then
x^2 + (3.5*10^-4)x - 0.25*(3.5*10^-4) = 0
solve for x
x =0.00918
substitute
[H+] = 0 + 0.00918= 0.01434 M
pH = -log(H+) = -log(0.00918) = 2.04
choose D
Q18
This is a buffer so
initially
mmol of HF = 0.25*40 = 10
mmol of OH- = MV = 10*0.2 = 2
after reaction
mmol of HF left = 10-2 = 8
mmol o fF- formed = 0+2 = 2
pH =pKa + log(F-/HF)
p´Ka = -log(3.5*10^-4) = 3.455
pH= 3.455+ log(2/8)
pH = 2.85
choose C
Q19
half way
pH = pKa + log(F-/HF) is true
F- = HF by definition, since 50% has been converted to F-
pH = 3.45
choose C
Q20
in equivalence point...
expect hydrolysis
F-(aq) + H2O(l) <->HF+ OH-(aq)
Let HA --> HFand A- = F- for simplicity
since A- is formed
the next equilibrium is formed, the conjugate acid and water
A- + H2O <-> HA + OH-
The equilibrium s best described by Kb, the base constant
Kb by definition since it is an base:
Kb = [HA ][OH-]/[A-]
Ka can be calculated as follows:
Kb = Kw/Ka = (10^-14)/(3.5*10^-4) = 2.86*10^-11
get ICE table:
Initially
[OH-] = 0
[HA] = 0
[A-] = M
the Change
[OH-] = + x
[HA] = + x
[A-] = - x
in Equilibrium
[OH-] = 0 + x
[HA] = 0 + x
[A-] = M - x
substitute in Kb expression
Kb = [HA ][OH-]/[A-]
M = 0.111 in equivalenc epoint
2.86*10^-11 = x*x/(0.11-x)
solve for x
x^2 + Kb*x - M*Kb = 0
solve for x with quadratic equation
x = OH- =1.78*10^-6
[OH-] 1.78*10^-6
pOH = -log(OH-) = -log(1.78*10^-6= 5.75
pH = 14-5.75= 8.25
pH = 8.25
choose A
Q21
mmol of base = MV = 80*0.2 = 16
mmol of acid = 0.25*40 = 10
mmol o fOH- left = 16-10 = 6
[OH-] = 6/(40+80) = 0.05
pH = 14 + log(0.05) = 12.698
choose b
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