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) =

you need to take a look at the periodic system in order to sort these elements into pairs that would most likely exhibit similar chemical properties. first, you need to find elements that are found in the same group (groups are vertical in the periodic system), and then to see whether they have the same valence number. if they meet these criteria, they will have similar properties.

the first pair: ne (neon) and kr (krypton), are noble gases found in group 8

the second pair: li (lithium) and k (potassium), found in group 1

the third pair: s (sulfur) and te (tellurium), found in group 6