Pyridine, c5h5n, is a weak base with kb = 1.5 × 10-9. what amount of pyridinium chloride, c5h5nhcl, must be added to 1.00 l of 0.250 m c5h5n, in order to produce a buffer having ph 5.00? assume no change in volume.

We should add 0.3785 moles of pyridinium chloride

Explanation:

Step 1: Data given

Kb = 1.5 * 10^-9

pKb = -log(1.5 * 10^-9) = 8.82

Molarity of C5H5N = 0.250 M

Volume of the buffer = 1.00 L

pH =5

pOH = 14 -5 = 9

Step 2: Calculate [C5H5NHCl]

pOH = pKb + log [C5H5NCl] /[C5H5N]

9 = 8.82 + log [C5H5NCl] /[C5H5N]

0.18 = log [C5H5NCl] /[C5H5N]

[C5H5NCl] /[C5H5N] = 1.514

[C5H5NCl] = 1.514 * 0.250

[C5H5NCl] = 0.3785 M

pOH = 8.82 + log (0.3785/0.25) = 9

If pOH = 9 then pH is 5

Step 3: Calculate number of moles of C5H5NHCL

moles = Molarity / volume

moles = 0.3785 M / 1 L

moles = 0.3785 moles

We should add 0.3785 moles of pyridinium chloride

i think so it’s will be b yes i believe its

the answer is d. 74.02