Nitric acid, HNO2 is a weak acid with Ka= 4.5 x 10-4. It dissociates according t

Question

Nitric acid, HNO2 is a weak acid with Ka= 4.5 x 10-4. It dissociates according to:

HNO2 (aq) <--> H+ (aq) + NO2- (aq)

a.) What is the pH of a buffer solution which is 0.80 M in NO2- and 0.40 M in HNO2?

b.) What will be the pH of the resulting solution when 2.0 mL of 0.5 M HCl are added to 50 mL of this (from part a) buffer solution?

c.) What will be the pH of the resulting solution when 2.0 mL of 0.5 M NaOH are added to 50 mL of this (from part a) buffer solution?

An answer accompanied by a complete, step by step solution would be greatly appreciated. Thanks

Explanation / Answer

a) pH = pKa + log[NO2-]/[HNO2] pH = 3.65 [H+] = 2.25 x 10^-4 M b.) net [H+] = (2.25 x 10^-4x0.05 + 0.002x0.5)/.052 M =1.94 x 10^-2M pH = 1.71 c) OH- = 0.002 x 0.5 = 0.001 moles H+ = 2.25 x 10^-4x0.05 = 1.125 x 10^-5 moles

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