Chemical Thermodynamics NEET 2014: Silver Carbonate Ksp

What is the answer to the Chemical Thermodynamics NEET 2014 silver carbonate question?

Option B is correct for the Chemical Thermodynamics NEET 2014 silver carbonate question. The decisive step is undoing the natural logarithm with an exponential, not a power of ten.

This Chemistry MCQ belongs to Chemical Thermodynamics. Its hard rating is this question bank’s classification, not a measured student failure rate.

At 25°C, silver carbonate dissolves according to:

A g_{2} C O_{3} (s) ⇌ 2 A g^{+} (a q) + C O_{3}^{2 -} (a q)

The standard Gibbs energy change per mole of reaction and the gas constant are:

Δ G^{\circ} = + 63.3 k J m o l^{- 1}, R = 8.314 J K^{- 1} m o l^{- 1}

Find the solubility product in water. The choices are:

Answer: B. The full derivation follows below.

Why is the equilibrium constant the solubility product here?

For the dissolution reaction as written, the equilibrium constant is the solubility product. Pure solid silver carbonate has unit activity, so it is omitted from the equilibrium expression. Under the usual dilute-solution treatment:

K_{s p} = [A g^{+}]^{2} [C O_{3}^{2 -}]

The square on silver-ion concentration comes from its coefficient in the balanced equation. The supplied standard Gibbs energy directly determines this equilibrium constant.

Follow the official solution by writing the general relation, then replacing the constant with the solubility product: $Δ G^{\circ} = - R T \ln K$ $Δ G^{\circ} = - R T \ln K_{s p}$

Divide both sides by the product of the gas constant and absolute temperature. Then move the minus sign:

\frac{Δ G^{\circ}}{R T} = - \ln K_{s p}

\ln K_{s p} = - \frac{Δ G^{\circ}}{R T}

The supplied quantity is the standard Gibbs energy change, not the actual Gibbs energy change at equilibrium. Do not replace it with zero: the actual Gibbs energy change becomes zero at equilibrium, but the standard value need not.

How does positive 63.3 kJ give option B?

Convert the energy to joules and temperature to kelvin, then substitute into the rearranged formula. The result is a negative natural logarithm; taking its exponential gives option B. Keep the energy unit consistent with the joule unit in the supplied gas constant.

First convert the given quantities:

Δ G^{\circ} = + 63.3 k J m o l^{- 1} = + 63, 300 J m o l^{- 1}

$25^{\circ} C \approx 298 K$ Use 298 K throughout, as in the official solution. The denominator is:

R T = 8.314 \times 298 = 2477.572 J m o l^{- 1}

Substitute into the rearranged relation:

\ln K_{s p} = - \frac{63300}{8.314 \times 298} \approx - 25.55

Undo the natural logarithm:

K_{s p} = e^{- 25.55} \approx 8.0 \times 10^{- 12}

Select option B. Check the order of magnitude using the equivalent base-ten form:

\log_{10} K_{s p} = \frac{- 25.55}{2.303} \approx - 11.09

K_{s p} \approx 10^{- 11} \times 10^{- 0.09} \approx 8.0 \times 10^{- 12}

This connects the logarithm to the power of ten printed in the options. It places the result just below one hundred-billionth.

The positive standard Gibbs energy also requires a constant below one:

Δ G^{\circ} > 0 ⟹ K_{s p} < 1

However, all four choices are below one, so the sign check alone cannot select B. You must check the magnitude.

The question bank’s expected solving time is 90 seconds. Treat that classification as a practice benchmark, not a measured average or a guarantee.

Why does using a power of ten lead towards option A?

A power of ten gives option A’s scale because it incorrectly treats the natural logarithm as a common logarithm. Start from the correctly calculated intermediate result: $\ln K_{s p} \approx - 25.55$

The incorrect operation is:

K_{s p} = 10^{- 25.55} \approx 2.8 \times 10^{- 26} (incorrect)

This reaches option A’s erroneous order of magnitude, but does not reproduce A exactly. Using a coarser intermediate value in the same wrong operation gives:

10^{- 25.5} \approx 3.2 \times 10^{- 26} (incorrect)

That matches option A. The official intermediate value does not give that exact number.

The method error is changing the logarithm’s base without the conversion factor, not merely an arithmetic slip. Both of these forms are valid: $Δ G^{\circ} = - R T \ln K$

Δ G^{\circ} = - 2.303 R T \log_{10} K

Pick one form and keep its base throughout. For this question, use the natural-log form because it follows the official solution directly.

The prevention rule: undo a natural logarithm with an exponential; undo a common logarithm with a power of ten. $\ln K = x \Rightarrow K = e^{x}$ $\log_{10} K = x \Rightarrow K = 10^{x}$

How do you solve three related Chemical Thermodynamics questions?

Use the same Gibbs energy relation while tracking how the reaction changes. Reversing the reaction changes the energy’s sign and inverts the constant; doubling the equation doubles the energy and squares the constant. These are original practice variations based on the Chemical Thermodynamics NEET 2014 question, not additional NEET PYQs.

What are the standard Gibbs energy and constant for the reverse reaction?

The standard Gibbs energy changes sign, and the equilibrium constant becomes the reciprocal of the dissolution constant. At the same temperature:

2 A g^{+} (a q) + C O_{3}^{2 -} (a q) ⇌ A g_{2} C O_{3} (s)

Δ G_{r e v e r s e}^{\circ} = - 63.3 k J m o l^{- 1}

K_{r e v e r s e} = \frac{1}{K_{s p}} \approx \frac{1}{8.0 \times 10^{- 12}} = 1.25 \times 10^{11}

This is the reverse reaction’s constant, not its solubility product.

What changes if the entire dissolution equation is doubled?

The standard Gibbs energy doubles because it refers to the reaction as written. The logarithm doubles, so the equilibrium constant is squared:

Δ G_{n e w}^{\circ} = + 126.6 k J m o l^{- 1}

\ln K_{n e w} = 2 \ln K_{s p}

K_{n e w} = K_{s p}^{2} \approx (8.0 \times 10^{- 12})^{2} = 6.4 \times 10^{- 23}

Do not double the constant itself.

Must the standard Gibbs energy be zero at equilibrium?

No. The actual Gibbs energy change is zero at equilibrium, not necessarily the standard value. Start from: $Δ G = Δ G^{\circ} + R T \ln Q$

At equilibrium:

Q = K, Δ G = 0

Therefore:

0 = Δ G^{\circ} + R T \ln K \Rightarrow Δ G^{\circ} = - R T \ln K

The standard Gibbs energy is zero only when: $K = 1$

On your next timed attempt, write the energy conversion and the inverse-log step explicitly. Check both before selecting option B.

Next step: the past-paper archive on NEET JEEnius AI and search past NEET papers by year, subject or chapter, each with a worked solution (100 free searches a month).

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Frequently asked questions

What is the answer to the NEET 2014 silver carbonate question?

Option B is correct: Ksp is approximately 8.0 × 10⁻¹². Using ΔG° = −RT ln Ksp with ΔG° = 63,300 J mol⁻¹ and T = 298 K gives ln Ksp ≈ −25.55. Taking the exponential gives Ksp = e⁻²⁵·⁵⁵, not 10⁻²⁵·⁵⁵.

Why does using 10 instead of e give the wrong Ksp?

The relation ΔG° = −RT ln K uses a natural logarithm, so its inverse is an exponential with base e. Using 10⁻²⁵·⁵⁵ instead gives approximately 2.8 × 10⁻²⁶, which is the wrong order of magnitude. To use a common logarithm correctly, write ΔG° = −2.303RT log10 K.

Why is solid silver carbonate not included in the Ksp expression?

Pure solid silver carbonate has unit activity, so it is omitted from the equilibrium expression. Under the usual dilute-solution treatment, Ksp = [Ag⁺]²[CO₃²⁻]. The silver-ion concentration is squared because its coefficient in the balanced dissolution equation is two.

Is standard Gibbs energy zero at equilibrium?

No: the actual Gibbs energy change, ΔG, is zero at equilibrium, but the standard Gibbs energy change, ΔG°, need not be. At equilibrium, ΔG° = −RT ln K. Therefore, ΔG° is zero only when K = 1.