Excercise 16.67: determine the pH of an HNO2 solution of each of the following c

Question

Excercise 16.67: determine the pH of an HNO2 solution of each of the following concentrations. Part A: 0.520 M Part B: 0.110 M Part C: 1.5 x 10^-2 M Part D: in which cases can you NOT make the simplifying assumption that x is small? Excercise 16.67: determine the pH of an HNO2 solution of each of the following concentrations. Part A: 0.520 M Part B: 0.110 M Part C: 1.5 x 10^-2 M Part D: in which cases can you NOT make the simplifying assumption that x is small? Part A: 0.520 M Part B: 0.110 M Part C: 1.5 x 10^-2 M Part D: in which cases can you NOT make the simplifying assumption that x is small?

Explanation / Answer

Part-A

Concentration of HNO2 = 0.520M

Ka of HNO2 = 4.5x10^-4

for weak acids

[H+] = square root of KaxC

[H+= square root of (4.5x10^-4 x0.520)

[H+}= 1.529x10^-2M

-log[H+]= -log(1.529x10^-2)

PH= 1.82.

second method

HNO2 ------------------ H+ + NO2-

0.520                       0       0

-x                          +x        +x

0.520-x                 +x       +x

Ka= [H+][NO2-]/[HNO2]

4.5x10^-4= x*x/(0.520-x)

for solving the equation

x= 0.0151

[H+] = 0.0151M= 1.51x10^-2M

-log[H+]= -log[1.51x10^-2]

PH= 1.82.

so both the methods answer is same

Part-b

Concentration of HNO2= 0.110M

[H+] = square root of (4.5x10^-4x0.110)

[H+] = 0.704x10^-2M

-log[H+]= -log{0.704x10^-2]

PH=2.15

Part-c

concentration of HNO2= 1.5x10^-2M

[H+] = square root of (4.5x10^-4x1.5x10^-2)

[H+] = 2.598 x10^-3M

-log{H+} = -log(2.598x10^-3)

PH= 2.58

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