What is the pH of a 0.1 M Ca(NO2)2? (Ka for HNO2 = 4.5 x 10^-4). I solved the pr

Question

What is the pH of a 0.1 M Ca(NO2)2? (Ka for HNO2 = 4.5 x 10^-4).

I solved the problem already. The pH is 8.3. However, I'm confused as to why I need to use the Kb value here instead of the already given Ka value.



And also, why does the Henderson Hasslebach equation use a Kb value even if not given? Finding Kb is not difficult, it's just Kw/Ka, but I have an exam tonight and I just want to know why I need to use the Kb value even if the question doesn't tell me to. I need a crystal clear explanation please. Thanks so much.

Explanation / Answer

Ca(NO2)2 + 2H2O ----> Ca2+ + 2OH- + 2HNO2

Now when HNO2 dissociates into H+ and NO2-, we solve the problem using the Ka values.

In this problem since NO2- gives HNO2, both the reactions are reverse.

If Ka for HNO2 is 4.5 x 10^-4

Kb for NO2- is = 10^-14 / 4.5 x 10^-4 = 2.222 x 10^-11

=> Kb for Ca(NO2)2 = 2.222 x 10^-11

Now we need to use this Kb value to find the pH or pOH.

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