Part 1. We’re going to titrate formic acid with the strong base, NaOH. There is
ID: 514936 • Letter: P
Question
Part 1. We’re going to titrate formic acid with the strong base, NaOH. There is initially 100. mL of 0.50 M formic acid and the concentration of NaOH is 1.0 M.
. 1) What is the initial pH of the formic acid solution?
2) What is the percent ionization under initial conditions?
3) After the addition of 10 mL of NaOH, what is the pH?
4) After the addition of 25 mL of NaOH, what is the pH?
5) What volume of NaOH is required to reach the equivalence point?
6) What is the pH at the equivalence point?
7) What is the pOH at the equivalence point?
8) If, instead of NaOH being added, 0.05 moles of HCl is added by __________% bubbling the gas through the solution. Assume that the volume has not changed. What is the percent dissociation of formic acid? Part 2. Examine the following compounds. Determine if they are acidic, basic, or neutral when dissolved in water. A) N(CH3)H3Br A)____________ B) Fe(NO3)3 B)____________ C) NaNO3 C)____________ D) K3PO4 D)____________ E) HClO2 E)____________ Part 3. Identify the labeled points in the following titration curve. A ___________________________ B ___________________________ C ___________________________ D ___________________________ E ___________________________ F ___________________________ G ___________________________ H ___________________________ I ___________________________ J ___________________________
Explanation / Answer
part 1)
1)
Ka = 1.8 x 10^-4
concentration C = 0.50 M
[H+] = sqrt (Ka x C)
= sqrt (1.8 x 10^-4 x 0.50 )
[H+] = 9.49 x 10^-3 M
pH = -log [H+] = -log (9.49 x 10^-3)
pH = 2.02
2)
% ionization = [H+] / C x 100
= (9.4868 x 10^-3 / 0.5) x 100
% ionization = 1.90 %
3) After the addition of 10 mL of NaOH, what is the pH?
millimoles of HF = 100 x 0.5 = 50
millimoles of NaOH = 10 x 1 = 10
HF + NaOH -------------> F- + H2O
50 10 0 0
40 0 10
pH = pKa + log [salt / acid]
= 3.74 + log [10 / 40]
= 3.15
pH = 3.15
4)
this is half equivalence point :
pH = pKa
pH = 3.75
5)
volume = 50 mL