Answer to Question #147535 in General Chemistry for gia

Question

At 28°C and 0.982 atm, gaseous compound HA has density of 1.16 g/L. A quantity of 2.03 g of this compound is dissolved in water and diluted to exactly 1L. If the pH of the solution is 5.22 (due to ionization of HA) at 25°C, calculate Ka of the acid.

Answer 1

T=28°C= 301K, P=0.982atm,

Density of the gaseous compound=1.16g/L

Mass of the gaseous compound=2.03g

Volume occupied by gaseous compound= mass/density = 2.03/1.16=1.75L

Let's find the moles of the gaseous compound, using the ideal gas equation.

PV=nRT

n=PV/RT

n=(0.982x1.75)/(0.0821x301)

n= 0.07mol

Litres of the acidic solution= 1L

Molarity of the acidic solution [HA]= moles of dissolved gas/litres of solution

Molarity of acidic solution [HA]= 0.07mol/1L

[HA]=0.07M

Now, let's find the concentration of H⁺ if pH= 5.22

pH=-log[H⁺]

[H⁺]= 10^-5.22

[H⁺]= 6.03 x 10^-6M

The ionization of the weak acid HA is given by the equation

HA:::::::: H⁺ + A

The equilibrium concentrations can then be expressed as:

[HA]=0.07M - 6.03x10^-6M ~0.07M

[H⁺]= 6.03x 10^-6M

[A^-]= 6.03x10^-6M

Ka= [H⁺][A^-]/[HA]

K_a= (6.03x10^-6)²/0.07

K_a= 5.19x10^-10