Determine what mass of sodium nitrite, NaNO2, would be required to prepare a buffer, Buffer A, with a pH of 3.13 from 50.0 mL of 1.0 M nitrous acid, NaNO2.The Ka and pKa for nitrous acid is 4.0 x 10-4 and 3.40 respectively.

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

a:

by the combination of an unlimited number of elements

6.4 cm (max) - 25 cm (min)