Question

If the kₐ of a monoprotic weak acid is 8.1 x 10⁻⁶, what is the ph of a 0.10 m solution of this acid? ph =

Answer 1

The pH of a 0.10 M solution of a weak acid with a Ka of 8.1 x 10−6 is calculated to be approximately 6.55.

Step-by-step explanation:

If the Ka of a monoprotic weak acid is 8.1 x 10−6, to find the pH of a 0.10 M solution of this acid, we can use the following approach:

1. Write the dissociation equation for the weak acid (HA → H+ + A−).
2. Set up an ICE (Initial, Change, Equilibrium) table.
3. Assume that the dissociation produces x moles/liter of H+, so [H+] and [A−] will both be x and [HA] will be 0.10 - x.
4. Write the expression for Ka: Ka = ([H+][A−])/[HA].
5. Substitute the equilibrium concentrations into the Ka expression to solve for x.
6. Calculate the pH: pH = −log([H+]).

Assuming x is small compared to 0.10 and can be neglected, the Ka expression simplifies to Ka = x²/0.10. Solving for x yields x approximately equal to the square root of (Ka × initial HA concentration). Substituting the given values, we get x ≈ √(8.1 x 10−6 × 0.10) which gives x, the [H+], as approximately 2.85 x 10−7 M. Thus, pH ≈ −log(2.85 x 10−7) = 6.55.