Answer:
1.85 g
Explanation:
The strategy here is to utilize the Henderson-Hasselbach equation
pH = pKa + log [A⁻] / [HA]
to calculate the ratio log [A⁻] / [HA], and from there to calculate the concentration [A⁻] and finally the mass of NaNO₂ from the number of moles assuming the final buffer volume is 50.0 mL ( that is the volume does not change by the addition of NaNO₂)
pH = pKa + log [NO₂⁻]/[HNO₂]
3.13 = 3.40 + log [NO₂⁻]/[HNO₂]
- 0.27 = log [NO₂⁻]/[HNO₂]
taking the inverse log function to both sides of this equation
0.54 = [NO₂⁻]/[HNO₂]
Now [HNO₂] = 1.0 M, therefore [NO₂⁻] = [NaNO₂] =
0.54 x 1.0 M = 0.54 M
from M = mol / L we get
mol = 0.54 mol/L x 0.050L = 0.027 mol
the molar mass of NaNO₂ is = 68.99 g / mol, so the mass of 0.027 mol is
0.027 mol x 68.99 g/mol = 1.85 g
