Answer:
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
