Question

Nitrous Acid (HNO₂) was added to 50 gallons of pure water until the pH of the solution dropped to 3.50. The Log Acidity Constant (pK_A) for HA is 3.15. How much acid (in grams) was added to the water?

Answer 1

HNO₂ is a weak acid which can be seen from the value of pKa = 3.15 as Ka = 10^-3.15

So, for weak acid we have -

HNO₂ -> H⁺ +NO₂^-

Ka = [H⁺][NO₂^-]/[HNO₂]

Initially lets say we had C concentration of HNO₂ and x amount of it got dissociated so,

At equilibirum we have Cx amount of H⁺, NO₂^- = Cx and HNO₂ = C(1-x)

so, Ka = C²x²/C(1-x)

New pH = 3.5 so, [H⁺] = 10^-3.5 = Cx

10^-3.15 = 10^-3.5*x/(1-x) solving for x we get x = 0.691

So, as Cx = 10^-3.5 we get C = 0.0004576M

The volume of pure water taken = 50 gallons = 50*3.785 L = 189.25 L

So, the number of moles of HNO₂ in this much amount of water will be C*Volume = 0.0866 moles

Weight of 0.0866 moles of HNO₂ = 4.07 g