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How would the calculated value of Ksp be affected by the accidental and unknowing collection of solid lead(II) chloride via pipet in the Gibb’s Free Energy experiment? This would not affect the calculated value of Ksp. The calculated value of Ksp would be too small. The calculated value of Ksp would be too big.

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Step-by-step explanation:

It is known that relation between Gibb's free energy and equilibrium or reaction quotient is as follows.


\Delta G = \Delta G^(o) + 2.303 RT logQ ..... (1)

At equilibrium,
\Delta G = 0. Also, Q = K this means that reaction is at equilibrium.

Therefore, formula in equation (1) will become as follows.


\Delta G^(o) = -2.303 RT log K ....... (2)

It is known that for a sparingly soluble salt, K =
K_(sp)

Thus, equation (2) will become as follows.


\Delta G^(o) = -2.303 RT log K_(sp) ....... (3)

So, when value of
K_(sp) increases then there will be decrease in the value of
\Delta G^(o) according to equation (3).

Hence, we can conclude that the calculated value of
K_sp would be too big.

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