PART 1. A buffer solution is made that is 0.314 M in HNO 2 and 0.314 M in KNO 2

Question

PART 1.

A buffer solution is made that is 0.314 M in HNO2and 0.314 M in KNO2.

(1) If Ka for HNO2is 4.50×10-4, what is the pH of the buffer solution? __________

(2) Write the net ionic equation for the reaction that occurs when 0.094 mol HBr is added to 1.00 L of the buffer solution.

Use H3O+instead of H+

_________ + _________ ===> __________ + ____________

PART 2

A buffer solution is made that is 0.374 M in H2C2O4 and 0.374 M in NaHC2O4.

(1) If Ka for H2C2O4 is 5.90×10-2, what is the pH of the buffer solution? ____________

(2) Write the net ionic equation for the reaction that occurs when 0.077 mol KOH is added to 1.00 L of the buffer solution.

________ + ________ ===> ________ + ________

Explanation / Answer

Part 1 ka of HNO2 = 4.5x10-4  or pKa = 3.3468

given [HNO2] = {NO2-] =0.314M

the pH of buffer is given by Hendersen equation

pH -= pka + log [conjugate base]/[acid]

= 3.3468 + log 0.314/0.314

= 3.3468

2) HNO2 -----------> NO2- + H3O+

0.314 0.314 - initial

- - 0.094

0.408 0.22 - after

thus the net ionic reaction is

NO2- + H3O+ <.---------> HNO2 + H2O

Part 2

1) [acid]= [conjugate base ]= 0.374 M

and ka = 5.9x10-2 and pka = 1.229

Thus pH = pKa = 1.229

2) H2A +KOH ----------> HA- + H2O

0.374 - 0.374

- 0.077 -

0.297 0 0.451 -

Thus the netionic reaction is

H2 A + OH- -------> HA- + H2o

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