Question

3.5g unknown monoprotic acid with a Ka of 4.9x 10 ^-5 is dissolved in enough water to make a 100mL solution. If the pH of this solution is 2.45, what is the molar mass of the acid?

Answer 1

alternate fromula for finding out the week acid PH is

pH = 1/2 (pKa-logC)

where C is the concentration

find out the pKa value using ka

pKa = -log[Ka] = -log[4.9x 10 ^-5] = 4.3

pH is given 2.45

put all these in the above equation

2.45 = 1/2 (4.3 -logC)

(4.3 -logC)= 4.9

log C = 4.3-4.9

logC = -0.6

C = 10^-0.6

C = 0.2512 M

now we have the Molarity, weight, volume we can find out the molar mass using molarity formul

Molarity M = weight / molar mass x 1000 / volume in mL

0.2512 = (3.5 / molar mass) x 1000 / 100

molar mass = 35 / 0.2512

molar mass = 139.33 gram /mol