A 25.0- mL sample of 0.150 M hydrazoic acid is titrated with a 0.150 M NaOH solution. What is the pH after 13.3 mL of base is added? The K_aof hydrazoic acid is 1.9 × 10^-⁵.

Question

Quizplus · Accepted Answer

1. Write the balanced chemical equation:
HN3 + NaOH -> NaN3 + H2O

2. Calculate the number of moles of HN3 initially present:
moles HN3 = (0.150 mol/L) x (25.0 mL/1000 mL) = 0.00375 mol

3. Calculate the number of moles of NaOH added:
moles NaOH = (0.150 mol/L) x (13.3 mL/1000 mL) = 0.001995 mol

4. Determine the limiting reagent:
HN3 : NaOH = 1 : 1
Since the number of moles of NaOH added is less than the number of moles of HN3 initially present, HN3 is the limiting reagent, and the remaining moles of HN3 can be calculated as follows:

moles HN3 remaining = 0.00375 mol - 0.001995 mol = 0.001755 mol

5. Calculate the concentration of HN3 after the addition of NaOH:
volume HN3 remaining = (25.0 mL + 13.3 mL) / 1000 mL = 0.0383 L
[Moles HN3] = 0.001755 mol / 0.0383 L = 0.0458 M

6. Calculate the pKa of HN3: 
pKa = -log(Ka) = -log(1.9 x 10^-5) = 4.72

7. Calculate the pH 
Since HN3 is a weak acid, we will use an ICE table to solve for [H+] and pH:

HN3 + H2O -> H3O+ + N3-

I 0.0458 M    0          0           
C -x          x          x           
E 0.0458-x    x          x

Ka = (H3O+)(N3-)/(HN3) = x^2/(0.0458-x)
Using the quadratic formula gives x = [H3O+] = 1.18 x 10^-5 M
pH = -log[H3O+] = 4.78

Thus, the pH after 13.3 mL of NaOH is added is B) 4.78

A 250- ML Sample of 0