To calculate pH of starting NaNO2 solution and after each addition of HCl:
a) NaNO2 dissociates into ions: Na+ + NO2- and reacts with water:
Step 1: ICE Chart
The Kb expression for NO2- is:
Assuming the reaction is at 25°C: Kw = 1.0x10-14
Kb = 2.17 x 10-11
Step 2: Solve for x = [|OH-]
we can ignore x in the denominator
x = [OH-] = 1.47 x 10-6 M
Step 3: Calculate pOH:
pOH = 5.83
Knowing that NaNO2 is the salt of a strong base and weak acid calculate :
a: pH of 0.100 M NaNO2 solution. Ka of HNO2 = 4.6 x 10-4.
b: pH of the solution obtained on adding 20.0 ml of a 0.100 M HCl solution to 60.0 ml of a 0.100 M NaNO2 .
c: pH of the solution obtained on adding 120.0 ml of a 0.100 M HCl solution to 60.0 ml of a 0.100 M NaNO2
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Based on our data, we think this problem is relevant for Professor Hjorth-Gustin's class at San Diego Mesa College.