A 15.0 mL sample of 0.50 M HCl is added to a 100. mL sample of 0.478 M HNO2 (Ka = 4.0 x 10-4). What is the concentration of NO2- ions at equilibrium?

Answer 1

moles H⁺ from HCl = 0.015 L x 0.50 mol/L = 0.0075 moles H⁺

Final volume of mixture = 15.0 ml + 100. ml = 115 ml = 0.115 L

[HNO₂] = (100 ml)(0.478 M) = (115 ml)(x M) and x = 0.416 M HNO₂

[H+] = 0.0075 mol / 0.115 L = 0.0652 M (we can use this and assume H⁺ from HNO₂ is small)

HNO₂ <==> H⁺ + NO₂^-

Ka = [H⁺][NO₂^-] / [HNO₂]

4.0x10^-4 = (0.0652)(NO₂^-) / 0.416 -x (again, assume x is small relative to 0.416 so ignore it)

[NO₂^-] = (4.0x10^-4)(0.416) / 0.0652

[NO₂^-] = 2.55x10^-3 M

Compare this [NO₂^-] to what it would have been if HCl weren't added:

4.0x10^-4 = (x)(x) / 0.478 - x (again, assume x is small and ignore it in denominator)

x² = 1.9x10^-4

x = 1.4x10^-2 M which is almost 5x greater than when HCl is present. HCl reduces the dissociation of HNO2 based on Le Chatelier's Principle