Calculate the Ka of a 0.35M weak acid with a pH of 4.2.
1 answer:
Answer:
Ka = 1.14x10⁻⁸
Explanation:
First we <u>calculate [H⁺] from the pH</u>:
- pH = -log[H⁺]
- [H⁺] =
- [H⁺] = 6.31x10⁻⁵ M
For a monoprotic weak acid, the molar concentration of H⁺ of a solution can be expressed as:
- [H⁺] = √(C*Ka)
Where C is the molar concentration of the weak acid solution.
- 6.31x10⁻⁵ M =
- 1.14x10⁻⁸ = Ka
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The net ionic equation of chromium (iii) hydroxide reacting with nitrous acid:
Cr(OH)₃ (s) + 3H⁺ (aq) ⇒ Cr³⁺ (aq) + 3H₂O (l)
<h3>Further explanation</h3>The electrolyte in the solution produces ions.
The equation of a chemical reaction can be expressed in the equation of the ions
For strong electrolytes (the ionization rate = 1) is written in the form of separate ions, while the weak electrolyte (degree of ionization <1) is still written as an un-ionized molecule
In the ion equation, there is a spectator ion that is the ion which does not change (does not react) because it is present before and after the reaction
When these ions are removed, the ionic equation is called the net ionic equation
For gases and solids including water (H₂O) can be written as an ionized molecule
So only the dissolved compound is ionized ((expressed in symbol aq)
From the problem, it is stated that chromium (iii) hydroxide reacting with nitrous acid.
Reactions that occur:
Cr(OH)₃ (s) + 3HNO₂ (aq) ⇒ Cr(NO₂)₃ (aq) + 3H₂O (l)
Cr (OH)₃ solid form that does not decompose in the form of ions, as well as water (H₂O)
So the ionic equation becomes:
Cr(OH)₃ (s) + 3H⁺ (aq) + 3NO₂⁻ (aq) ⇒ Cr³⁺ (aq) + 3NO₂⁻ (aq) + 3H₂O (l)
There is an ion spectator that is 3NO₂⁻, so that if it is removed a net ionic equation will be formed:
Cr(OH)₃ (s) + 3H⁺ (aq) ⇒ Cr³⁺ (aq) + 3H₂O (l)
<h3>Learn more</h3>the net ionic equation
Keywords: the net ionic equation, an ion spectator
