How many grams of the bromide salt of the conjugate acid must be combined with how many grams of the weak base, to produce 1.00 L of a buffer that is

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) =

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