he pKa of HNO2 is 3.37. The pH of a 0.01 M aqueous solution of HNO2 is

To find the pH of a 0.01 M aqueous solution of HNO2, we can use the formula for the ionization of a weak acid and the definition of pKa.

1. Write the ionization equation for HNO2: HNO2 ⇌ H+ + NO2-

2. Write the expression for the acid dissociation constant (Ka): Ka = [H+][NO2-] / [HNO2]

3. Convert the pKa to Ka: Ka = 10^-pKa = 10^-3.37 = 4.27 x 10^-4

4. Set up an ICE (Initial, Change, Equilibrium) table to find the concentrations of H+, NO2-, and HNO2 at equilibrium.

   Initial: [HNO2] = 0.01 M, [H+] = [NO2-] = 0 Change: [HNO2] = -x, [H+] = [NO2-] = +x Equilibrium: [HNO2] = 0.01 - x, [H+] = [NO2-] = x

5. Substitute the equilibrium concentrations into the Ka expression and solve for x: 4.27 x 10^-4 = x^2 / (0.01 - x)

Assuming x is much smaller than 0.01 (which is a valid assumption for weak acids), we can simplify this to: 4.27 x 10^-4 = x^2 / 0.01 x = sqrt(4.27 x 10^-4 * 0.01) = 0.0020656 M

6. The concentration of H+ ions, [H+], is equal to x, so the pH of the solution is: pH = -log[H+] = -log(0.0020656) = 2.68

So, the pH of a 0.01 M aqueous solution of HNO2 is approximately 2.68.