Answer:
pH =3.8
Explanation:
Lets call the monoprotic weak acid HA, the dissociation equilibria in water will be:
HA + H₂O ⇄ H₃O⁺ + A⁻ with Ka = [ H₃O⁺] x [A⁻]/ [HA]
The pH is the negative log of the H₃O⁺ concentration, we know the equilibrium constant, Ka and the original acid concentration. So we will need to find the [H₃O⁺] to solve this question.
In order to do that lets set up the ICE table helper which accounts for the species at equilibrium:
HA H₃O⁺ A⁻
Initial, M 0.40 0 0
Change , M -x +x +x
Equilibrium, M 0.40 - x x x
Lets express these concentrations in terms of the equilibrium constant:
Ka = x² / (0.40 - x )
Now the equilibrium constant is so small ( very little dissociation of HA ) that is safe to approximate 0.40 - x to 0.40,
7.3 x 10⁻⁶ = x² / 0.40 ⇒ x = √( 7.3 x 10⁻⁶ x 0.40 ) = 1.71 x 10⁻³
[H₃O⁺] = 1.71 x 10⁻³
Indeed 1.71 x 10⁻³ is small compared to 0.40 (0.4 %). To be a good approximation our value should be less or equal to 5 %.
pH = - log ( 1.71 x 10⁻³ ) = 3.8
Note: when the aprroximation is greater than 5 % we will need to solve the resulting quadratic equation.
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HF and NaF - If the right concentrations of aqueous solutions are present, they can produce a buffer solution.
<h3>What are buffer solutions and how do they differ?</h3>
- The two main categories of buffers are acidic buffer solutions and alkaline buffer solutions.
- Acidic buffers are solutions that contain a weak acid and one of its salts and have a pH below 7.
- For instance, a buffer solution with a pH of roughly 4.75 is made of acetic acid and sodium acetate.
<h3>Describe buffer solution via an example.</h3>
- When a weak acid or a weak base is applied in modest amounts, buffer solutions withstand the pH shift.
- A buffer made of a weak acid and its salt is an example.
- It is a solution of acetic acid and sodium acetate CH3COOH + CH3COONa.
learn more about buffer solutions here
<u>brainly.com/question/8676275</u>
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sorry i don't know the answer i'm really sorry
