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

User Oersted
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1 Answer

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Answer:

[H⁺] = 3.23x10⁻³ M

Step-by-step 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
User Hammer
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