A chemistry graduate student is given 125.mL of a 1.00M benzoic acid HC6H5CO2 solution. Benzoic acid is a weak acid with =Ka×6.310−5.What mass of KC6H5CO2 should the student dissolve in the HC6H5CO2 solution to turn it into a buffer with pH =4.63? You may assume that the volume of the solution doesn't change when the KC6H5CO2 is dissolved in it

He should dissolve 53.9 grams of KC6H5CO2

Explanation:

Step 1: Data given

Volume of A 1.00 M benzoic acid solution = 125 mL = 0.125 L

Ka of benzoic acid = 6.3*10^-5

pH = 4.63

Step 2: Calculate concentration of conjugate base

pH = pKa + log ([A-]/[HA])

4.63 = 4.20 + log ([A-]/[HA])

0.43 = log ([A-]/[HA])

10^0.43 = [A-]/[HA])

2.69 = [A-]/[HA])

2.69 = [A-]/ 1.00

[A-] = 2.69 M

Step 3: Calculate moles KC6H5CO2

Moles molarity * volume

Moles = 2.69 M * 0.125 L

Moles = 0.33625 moles

Step 4: Calculate mass KC6H5CO2

Mass of KC6H5CO2 = moles * molar mass

Mass of KC6H5CO2 = 0.33625 moles * 160.21 g/mol

Mass of KC6H5CO2 = 53.9 grams

He should dissolve 53.9 grams of KC6H5CO2

c- cold periods in earth's history during which glaciers covered much of the surface.

explanation: