Answer:
79.1g of weak base must be combined with 56.0g of conjugate acid
Explanation:
that is 1.00 M in the weak base?
The weak base is C5H5N with a pKa of 5.17 (Ka=6.7×10-6)and a desire pH of 5.63
The equilibrium of the weak base is with the bromide salt of the conjugate acid is:
C5H5N(aq) + H2O(l) + Br- ⇄ C5H5NHBr(aq) + OH-(aq)
Where Kb = Kw / Ka = 1x10⁻¹⁴ / 6.7x10⁻⁶
Kb = 1.49x10⁻⁹ is defined as:
Kb = 1.49x10⁻⁹ = [C5H5NHBr] [OH-] / [C5H5N]
Where [OH-] = 10^-(14- pH) = 10^-(14- 5.63) = 4.255x10⁻⁹M
[C5H5N] = 1.00M
Replacing:
1.49x10⁻⁹ = [C5H5NHBr] [OH-] / [C5H5N]
1.49x10⁻⁹ = [C5H5NHBr] [4.255x10⁻⁹M] / [1.00M]
[C5H5NHBr] = 0.35M
In 1L the moles of C5H5NHBr are 0.35 moles
Mass C5H5NHBr: 160.0118g/mol
0.35 moles * (160.0118g / mol) =
56.0g of C5H5NHBr are necessaries
The mass of C5H5N is -79.1g/mol-:
1.00moles * (79.1g/mol) =
