1.4406 × 10⁻⁶ is the Ka(dissociation constant) for the acid.
The equilibrium constant for the reaction of an acid with water is the acid dissociation constant, where the acid, HA separates into H⁺ and A⁻ ions.
The acid dissociation constant is represented by (Ka).
So let's first imagine that the given monoprotic acid is HA.
HA will dissociate into H⁺ and A⁻ ions.
HA ⇒ H⁺ and A⁻
The formula used for Ka is
Ka = [H⁺] [A⁻] / [HA]
Given
pH = 2.83
[HA] = 1.50 M
From the given pH, we can calculate [H⁺] and [A⁻]
[H⁺] = [A⁻] = 1 × = 1.47 × 10⁻³ M
Ka = 1.47 × 10⁻³ × 1.47 × 10⁻³ / 1.50
Ka = 1.4406 × 10⁻⁶
Hence, 1.4406 × 10⁻⁶ is the Ka for the acid.
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Answer:
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Explanation:
Answer: 240 mL
Explanation:
(20 kPa)(300 mL) = (25 kPa)V2
V2 = 240 mL
Or my way which was
300 mL / 25 kPa =12
12(5)=60
300-60=240mL
Products over reactants:
<span>HS- + H2O ----> H2S + OH-
</span><span>The expression represents Kb for HS–</span>