Respuesta :
Answer:
[OH⁻] = 4.3 x 10⁻¹¹M in OH⁻ ions.
Explanation:
Assuming the source of the carbonate ion is from a Group IA carbonate salt (e.g.; Na₂CO₃), then 0.115M Na₂CO₃(aq) => 2(0.115)M Na⁺(aq) + 0.115M CO₃²⁻(aq). The 0.115M CO₃²⁻ then reacts with water to give 0.115M carbonic acid; H₂CO₃(aq) in equilibrium with H⁺(aq) and HCO₃⁻(aq) as the 1st ionization step.
Analysis:
H₂CO₃(aq) ⇄ H⁺(aq) + HCO₃⁻(aq); Ka(1) = 4.3 x 10⁻⁷
C(i) 0.115M 0 0
ΔC -x +x +x
C(eq) 0.115M - x x x
≅ 0.115M
Ka(1) = [H⁺(aq)][HCO₃⁻(aq)]/[H₂CO₃(aq)] = [(x)(x)/(0.115)]M = [x²/0.115]M
= 4.3 x 10⁻⁷ => x = [H⁺(aq)]₁ = SqrRt(4.3 x 10⁻⁷ · 0.115)M = 2.32 x 10⁻⁴M in H⁺ ions.
In general, it is assumed that all of the hydronium ion comes from the 1st ionization step as adding 10⁻¹¹ to 10⁻⁷ would be an insignificant change in H⁺ ion concentration. Therefore, using 2.32 x 10⁻⁴M in H⁺ ion concentration, the hydroxide ion concentration is then calculated from
[H⁺][OH⁻] = Kw => [OH⁻] = (1 x 10⁻¹⁴/2.32 x 10⁻⁴)M = 4.3 x 10⁻¹¹M in OH⁻ ions.
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NOTE: The 2.32 x 10⁻⁴M value for [H⁺] is reasonable for carbonic acid solution with pH ≅ 3.5 - 4.0.
The concentration of hydroxide ion of given solution is 4.3 x 10⁻¹¹M.
How we calculate the [OH⁻]?
We can calculate the concentration of hydroxide ions as follow:
[OH⁻][H⁺] = 10⁻¹⁴
Given chemical reaction with ICE table shown as below:
H₂CO₃(aq) ⇄ H⁺(aq) + HCO₃⁻(aq)
Initial: 0.115 0 0
Change: -x +x +x
Equilibrium: 0.115-x +x +x
Given that, Ka = 4.3 x 10⁻⁷
Equilibrium constant for this reaction is written as:
Ka = [H⁺][HCO₃⁻]/[H₂CO₃]
4.3 x 10⁻⁷ = x² / 0.115
x = 2.32 x 10⁻⁴M = [H⁺]
Now we calculate the concentration of hydroxide ion as:
[OH⁻][H⁺] = 10⁻¹⁴
[OH⁻] = 10⁻¹⁴ / 2.32 x 10⁻⁴ = 4.3 x 10⁻¹¹M
Hence, value of [OH⁻] is 4.3 x 10⁻¹¹M.
To know more about pH & pOH, visit the below ink:
https://brainly.com/question/24595796
