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What are the concentrations of [K+], [OH-], [CO32-] and [H+], in a 1.2 M solution of K2CO3 ? (Note: Question is asking for concentrations and not pH) g

User Rus
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Answer:

The concentrations are: [K⁺] = 1.2 M, [OH⁻] = 0.016 M, [CO₃²⁻] = 1.18 M and [H⁺] = 6.25x10⁻¹³ M.

Step-by-step explanation:

The dissociation equation of K₂CO₃ in water is:

K₂CO₃(aq) ⇄ K⁺(aq) + CO₃²⁻(aq) (1)

Also, the CO₃²⁻ will react with water as follows:

CO₃²⁻(aq) + H₂O(l) ⇄ HCO₃⁻(aq) + OH⁻(aq) (2)

The constant of the reaction (2) is:


Kb = ([OH^(-)][HCO_(3)^(-)])/([CO_(3)^(-2)]) = 2.08 \cdot 10^(-4)

The solution of K₂CO₃ is 1.2 M, and since the mole ratio of K₂CO₃ with K⁺ and CO₃²⁻ is 1:1, then we have:


[K_(2)CO_(3)] = [K^(+)] = [CO_(3)^(-2)] = 1.2 M

Now, from equation (2) we have:

CO₃²⁻(aq) + H₂O(l) ⇄ HCO₃⁻(aq) + OH⁻(aq) (3)

1.2 - x x x


2.08 \cdot 10^(-4) = ([OH^(-)][HCO_(3)^(-)])/([CO_(3)^(-2)])


2.08 \cdot 10^(-4) = (x^(2))/(1.2 - x)


2.08 \cdot 10^(-4)*(1.2 - x) - x^(2) = 0 (4)

By solving equation (4) for x we have:

x = 0.016 M = [HCO₃⁻] = [OH⁻]

Hence, the CO₃²⁻ concentration is:

[CO₃²⁻] = 1.2 M - 0.016 M = 1.18 M

Finally, the concentration of [H⁺] is:


[H^(+)][OH^(-)] = 10^(-14)


[H^(+)] = (10^(-14))/([OH^(-)]) = (10^(-14))/(0.016) = 6.25 \cdot 10^(-13) M

Therefore, the concentrations are: [K⁺] = 1.2 M, [OH⁻] = 0.016 M, [CO₃²⁻] = 1.18 M and [H⁺] = 6.25x10⁻¹³ M.

I hope it helps you!

User Fzerorubigd
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