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

the electron configuration that you gave me has 16 electrons ergo it has 16 protons ergo it is the element sulfur.