Question

Part A

A 50.0-mL sample of 0.200 M sodium hydroxide is titrated with 0.200 M nitric acid.

Calculate the pH of the solution, after you add a total of 51.9 mL 0.200 M HNO3.

Express your answer using two decimal places.

Part B

A 39.0 mL sample of 0.146 M HNO2 is titrated with 0.300 M KOH. (Ka for HNO2 is 4.57×10⁻⁴.)

Determine the pH at the equivalence point for the titration of HNO2 and KOH

.

Answer 1

A)

Given:

M(HNO3) = 0.2 M

V(HNO3) = 51.9 mL

M(NaOH) = 0.2 M

V(NaOH) = 50 mL

mol(HNO3) = M(HNO3) * V(HNO3)

mol(HNO3) = 0.2 M * 51.9 mL = 10.38 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.2 M * 50 mL = 10 mmol

We have:

mol(HNO3) = 10.38 mmol

mol(NaOH) = 10 mmol

10 mmol of both will react

remaining mol of HNO3 = 0.38 mmol

Total volume = 101.9 mL

[H+]= mol of acid remaining / volume

[H+] = 0.38 mmol/101.9 mL

= 3.729*10^-3 M

use:

pH = -log [H+]

= -log (3.729*10^-3)

= 2.4284

Answer: 2.43

B)

find the volume of KOH used to reach equivalence point

M(HNO2)*V(HNO2) =M(KOH)*V(KOH)

0.146 M *39.0 mL = 0.3M *V(KOH)

V(KOH) = 18.98 mL

Given:

M(HNO2) = 0.146 M

V(HNO2) = 39 mL

M(KOH) = 0.3 M

V(KOH) = 18.98 mL

mol(HNO2) = M(HNO2) * V(HNO2)

mol(HNO2) = 0.146 M * 39 mL = 5.694 mmol

mol(KOH) = M(KOH) * V(KOH)

mol(KOH) = 0.3 M * 18.98 mL = 5.694 mmol

We have:

mol(HNO2) = 5.694 mmol

mol(KOH) = 5.694 mmol

5.694 mmol of both will react to form NO2- and H2O

NO2- here is strong base

NO2- formed = 5.694 mmol

Volume of Solution = 39 + 18.98 = 57.98 mL

Kb of NO2- = Kw/Ka = 1*10^-14/4.57*10^-4 = 2.188*10^-11

concentration ofNO2-,c = 5.694 mmol/57.98 mL = 0.0982M

NO2- dissociates as

NO2- + H2O -----> HNO2 + OH-

0.0982 0 0

0.0982-x x x

Kb = [HNO2][OH-]/[NO2-]

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((2.188*10^-11)*9.821*10^-2) = 1.466*10^-6

since c is much greater than x, our assumption is correct

so, x = 1.466*10^-6 M

[OH-] = x = 1.466*10^-6 M

use:

pOH = -log [OH-]

= -log (1.466*10^-6)

= 5.8339

use:

PH = 14 - pOH

= 14 - 5.8339

= 8.1661

Answer: 8.17