2. (8 pts) Boric acid (H3BO3) has three hydrogens in a molecule, but effectively acts as a monoprotic acid (Ka = 5.8∙10-10), since the second and third Ka’s are negligible (less than 10-14). As any acid, H3BO3 will react with a strong base forming a salt (and possibly water). How many grams of boric acid and how many grams of NaOH are needed to prepare 1.00 L of a buffered solution with pH = 9.00 and a total concentration of boron 0.200 mol/L?

7.85 g H₃BO₃

2.92 g NaOH

Explanation:

The strategy for solving this question is to first utilize the Henderson- Hasselbach equation to calculate the ratio of conjugate base concentration to weak acid:

pH = pKa + log [A⁻]/ [HA]

In this case:

pH = pKa + log [H₂BO₃⁻]/[H₃BO₃]

We know pH and indirectly pKa ( = - log Ka ).

9.00 = -log(5.8 x 10⁻¹⁰) + log [H₂BO₃⁻]/[H₃BO₃]

9.00 = 9.24 + log [H₂BO₃⁻]/[H₃BO₃]

log [H₂BO₃⁻]/[H₃BO₃] = - 0.24

taking inverse log function to both sides of the equation:

[H₂BO₃⁻]/[H₃BO₃]  = 10^-0.24 = 0.58

We are also told we want to have a total concentration of boron of 0.200 mol/L, and if we call x the concentration of  H₂BO₃⁻ and y the concentration of H₃BO₃, it follows that:

x + y = 0.200 ( since we have 1 Boron atom per formula of each compound)

and from the Henderson Hasselbach calculation, we have that

x / y = 0.58

So we have a system of 2 equations with two unknowns, which when solved give us that

x = 0.073  and y = 0.127

Because we are told the volume is one liter it follows that the number of moles of boric acid and the salt are the same numbers 0.073 and 0.127

gram boric acid = 0.127 mol x molar mass HBO₃ = 0127 mol x 61.83 g/mol

                          = 7.85 g boric acid

grams NaOH = 0.073 mol x molar mass NaOH = 0.073 x 40 g/mol

                          = 2.92 g NaOH

93.2 ml + 3920 ml = 4013 ml total  

(93.2 ml) * (2.03 m) / (4013 ml) = 0.0471 m

the volume in in l of the diluted solution would be 0.0471 m.

2b2t

explanation:

2b2t