What mass of solid potassium nitrite would need to be dissolved in 500. mL of 0.40 M HNO2 to make 500. mL of a NO2−/HNO2 buffer solution that has a pH = 3.25? Ka of HNO2 = 7.1 x 10−4. (Assume no significant volume change during dissolution of the potassium nitrite)

Question

What mass of solid potassium nitrite would need to be dissolved in 500. mL of 0.40 M HNO2 to make 500. mL of a NO2−/HNO2 buffer solution that has a pH = 3.25?  Ka of HNO2 = 7.1 x 10−4.  (Assume no significant volume change during dissolution of the potassium nitrite)
	(1)  3.3 g     (2)  6.5 g     (3)  11 g     (4)  21 g     (5)  42 g

I know the answer is 21 g, but I cannot seem to get this answer when I'm working out the problem.

DrBob222 · Answer

500 mL x 0.40M = 200 millimols HNO2 = 0.200 mol HNO2 3.25 = 3.15 + log(base)/(.200) mols base = 0.252 0.252 x 85.1 = 21.4 g.