Question

At 28°C and 0.982 atm, gaseous compound HA has a density of 1.16 g/L. A quantity of 3.34 g of this compound is dissolved in water and diluted to exactly 1 L. If the pH of the solution is 5.01 (due to the ionization of HA) at 25°C, calculate the Ka of the acid. Please explain your steps.

Answer 1

First, identify molar mass

PV = nRT

PV = mass/MW * RT

MW = mass/V * RT/P

D = P*MW/(RT)

MW = DRT/P = (1.16)(0.082)(28+273)/(0.982)

MW = 29.155 g/mol

mol of sample = mass/MW = 3.34/29.155

mol = 0.114560

[HA]initially = mol of HA / Volume =  0.114560 / 1 = 0.114560 M

pH = 5.01

[H+] = 10^_pH = 10^-5.01 = 0.00000977237 M

then

HA <-> H+ + A-

Keq:

Ka = [H+][A-]/[HA]

[H+] = 0.00000977237

[A-] = [H+] = 0.00000977237

[HA] = 0.114560 - 0.00000977237 = 0.114550

substitute in Ka

KA = (0.00000977237*0.00000977237) / (0.114550)

Ka = 8.3369*10^-10