How many moles of HNO2 and NaNO2 remain in solution after addition of the HCl?

Question

1.00 mL of 12.0 M HCl is added to 1.00 L of a buffer that contains  0.130 M HNO₂ and 0.200 M NaNO₂. How many moles of HNO₂ and NaNO₂ remain in solution after addition of the HCl?

___ mol HNO₂ remain

____ mol NaNO₂ remain

Answer 1

Set up the reaction:

HCl (acid) will react with NO₂^- (base) in the buffer. This will form HNO₂ as follows:

H⁺ + NO₂^- == HNO₂

moles H⁺ added = 1.00 ml x 1 L / 1000 ml x 12.0 mol/L = 0.012 moles H⁺

H⁺ + NO₂^- == HNO₂

0.012....0.200.......0.130..........Initial

-0.012..-0.012.....+0.012.........Change

0...........0.188.....0.142.........Equilibrium (assuming no change in volume by addition of 1 ml HCl)

Use the Henderson Hasselbalch equation to solve for pH:

pH = pKa + log [NO₂^-]/[HNO₂] and pKa for HNO₂ = 3.39 (looked it up in a table)

pH = 3.39 + log (0.188 / 0.142) = 3.39 + 0.122

pH = 3.51