A certain monoprotic weak acid with Ka = 0.49 can be used in various industrial processes. (a) What is the [H+] for a 0.191 M aqueous solution of this acid and (b) what is its pH? Round the [H+] to three significant figures and the pH to two places past the decimal and do NOT use scientific notation
how would i do this?
To determine the [H+] and pH of a solution of a monoprotic weak acid with a given Ka value, you can follow these steps:
(a) To find the [H+] for a 0.191 M aqueous solution of this acid:
1. Write the balanced chemical equation for the dissociation of the weak acid in water.
Example: HA ⇌ H+ + A-
2. Set up an ICE table (Initial, Change, Equilibrium) for the dissociation reaction.
Initial: [HA] = 0.191 M, [H+] = 0 M, [A-] = 0 M
Change: -x, +x, +x
Equilibrium: 0.191 - x, x, x
3. Write the expression for Ka (acid dissociation constant).
Ka = [H+][A-] / [HA]
4. Substitute the equilibrium concentrations into the Ka expression.
Ka = x^2 / (0.191 - x)
5. Since x will be small compared to 0.191 (due to the weak nature of the acid), you can approximate 0.191 - x as approximately 0.191. This simplifies the equation to:
Ka = x^2 / 0.191
6. Solve for x by rearranging the equation:
x^2 = Ka * 0.191
x = √(Ka * 0.191)
7. Substitute the given Ka value into the equation:
x = √(0.49 * 0.191)
x = √0.09459
x ≈ 0.307
8. The [H+] concentration is equal to x.
[H+] = 0.307 M
(b) To find the pH of the solution, you can use the equation:
pH = -log[H+]
9. Substitute the [H+] concentration into the equation:
pH = -log(0.307)
pH ≈ 0.513
Therefore, the [H+] for the 0.191 M solution of this weak acid is 0.307 M, and its pH is approximately 0.513.