Question

What is [H⁺] in a 0.330 M solution of acrylic acid, CH₂CHCOOH (Ka = 3.16 × 10⁻⁵)?

Answer 1

Answer:

[H⁺] = 3.23x10⁻³ M

Explanation:

For a weak monoprotic acid, such as acrylic acid, we can calculate [H⁺] using the following formula if the solution concentration is significantly higher than the Ka:

• [H⁺] = √(C * Ka)

Where C is the weak acid concentration.

We input the data:

• [H⁺] = $\sqrt{0.330 M * 3.16x10^{-5}}$

And solve for [H⁺]:

• [H⁺] = 3.23x10⁻³ M