Question

Be sure to answer all parts. A 0.365−mol sample of HX is dissolved in enough H2O to form 835.0 mL of solution. If the pH of the solution is 3.70, what is the Ka of HX? Be sure to report your answer to the correct number of significant figures.

Answer 1

Answer:

The Ka is 9.11 *10^-8

Explanation:

Step 1: Data given

Moles of HX = 0.365

Volume of the solution = 835.0 mL = 0.835 L

pH of the solution = 3.70

Step 2: Calculate molarity of HX

Molarity HX = moles HX / volume solution

Molarity HX = 0.365 mol / 0.835 L

Molarity HX = 0.437 M

Step 3: ICE-chart

[H+] = [H3O+] = 10^-3.70 = 1.995 *10^-4

Initial concentration of HX = 0.437 M

Initial concentration of X- and H3O+ = 0M

Since the mole ratio is 1:1; there will react x M

The concentration at the equilibrium is:

[HX] = (0.437 - x)M

[X-] = x M

[H3O+] = 1.995*10^-4 M

Since 0+x = 1.995*10^-4   ⇒ x=1.995*10^-4

[HX] = 0.437 - 1.995*10^-4 ≈ 0.437 M

[X-] = x = 1.995*10^-4 M

Step 4: Calculate Ka

Ka = [X-]*[H3O+] / [HX]

Ka = ((1.995*10^-4)²)/ 0.437

Ka = 9.11 *10^-8

The Ka is 9.11 *10^-8