Please provide explanation and work for parts c, d, and e of this question. Find

Question

Please provide explanation and work for parts c, d, and e of this question.

Find the pH during the titration of 20.00 mL of 0.1250 M nitrous acid, HNO_2 (K_a = 7.1 times 10^-4), with 0.1250 M NaOH solution after the following additions of titrant. 0 mL 10.00 mL 15.00 mL Your response differs from the correct answer by more than 10%. Double check your calculations 20.00 mL Your response differs from the correct answer by more than 10%. Double check your calculations, 25.00 mL Your response differs from the correct answer by more than 10%. Double check your calculations.

Explanation / Answer

millimoles of HNO2 = 20 x 0.125 = 2.5

pKa = - log Ka = - log [7.4 x 10-4] = 3.15

c) millimoles of NaOH = 15 x 0.125 = 1.875

2.5 - 1.875 = 0.625 millimoles acid left

1.875 millimoles of salt formed

[HNO2] = [0.625] / 35 = 0.018 M

[NaNO2] = 1.875 / 35 = 0.053 M

pH = pKa + [NaNO2] / [HNO2]

pH = 3.15 + log [0.053] / [0.018]

pH = 3.62

d) millimoles of NaOH = 20 x 0.125 = 2.5

it is equivalence point

pH = 1/2 [pKw + pKa + log C]

C = 0.125 / 2 = 0.0625 M

pH = 1/2 [14 + 3.15 + log 0.0625]

pH = 1/2[15.94]

pH = 7.97

e) millimoles of NaOH = 25 x 0.125 = 3.125

3.125 - 2.5 = 0.625 millimoles of NaOH

[NaOH] = 0.625 / 45 = 0.014 M

pOH = - log [OH-]

pOH = - log [0.014]

pOH = 1.85

pH = 14 - 1.85

pH = 12.15

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.