Problem

What is the effect on the amount of CaHPO_ that d…

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Problem 112 Hard Difficulty

What is the effect on the amount of solid $\mathrm{Mg}(\mathrm{OH})_{2}$ that dissolves and the concentrations of $\mathrm{Mg}^{2+}$ and $\mathrm{OH}^{-}$ when each of the following are added to a mixture of solid $\mathrm{Mg}(\mathrm{OH})_{2}$ and water at equilibrium?
(a) $\mathrm{MgCl}_{2}$
(b) $\mathrm{KOH}$
(c) $\mathrm{HClO}_{4}$
(d) $\mathrm{NaNO}_{3}$
(e) $\mathrm{Mg}(\mathrm{OH})_{2}$

Answer

a. In quantitative terms, the added Mg"t causes the reaction quotent to be larger
Than the solubilly product $Q_{v}>K_{p}_{p}$ , and $\mathrm{Mg}(\mathrm{OH})_{2}$ forms untilithe reaction quotient again
equals $K_{p}$ At the new equillom, loh $^{-}$ listess and $\left[\mathrm{Mg}^{2+}\right]$ is greater than in the solution of
$\mathrm{Mg}(\mathrm{OH})_{2}$ in pure water.
More solid $\mathrm{Mg}(\mathrm{OH})_{2}$ is present
b. The reaction shifts to the left to minimize the stress produced by the additional $\mathrm{OH}^{-}$ ion.
$\mathrm{Mg}(\mathrm{OH})_{2}$ . Forms until the reaction quotient becomes equal to the solubility product.
At the new equilibrium, $\left[\mathrm{Mg}^{2+}\right]$ is less and $\left[\mathrm{OH}^{-}\right]$ is greater than in the solution of $\mathrm{Mg}(\mathrm{OH})_{2}$ , in
pure water.
More solid $\mathrm{Mg}(\mathrm{OH})_{2}$ , is present.
c. the decrease in the $\left[\mathrm{OH}^{-}\right]$ causes the reaction quotient to be smaller than the solubility product $\left(Q_{s p}<K_{s p}\right),$ and additional $\mathrm{Mg}(\mathrm{OH})_{2}$ dissolves until the reaction quotient again equals
$K_{s p}$ . At the new equilibrium, $\left[\mathrm{OH}^{-}\right]$ is less and $\left[\mathrm{Mg}^{2+}\right]$ is greater than in the solution of
$\mathrm{Mg}(\mathrm{OH})_{2}$ in pure water.
More $\mathrm{Mg}(\mathrm{OH})_{2}$ is dissolved.
d. NaNO_ $_{3}$ does not affect the solubility of $\mathrm{Mg}(\mathrm{OH})_{2}$
e. changing the amount of solid magnesium hydroxide in the mixture has no effect on the value of $Q,$ so there is no shifting of the reaction to restore $Q$ to the value of the equilibrium constant.

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Video Transcript

So let me start by writing the reaction for make Bijan I go excite for part A. The reaction shifts towards the left, so to the left, to minimize distress produced by additional magnesia poop or the divine according to really shat Leah's principle The Soul Ability Product on MG, which forms until the reaction caution again, is equal to K. S P. Move it along to option number me the reaction shift towards the left to minimize distress produced by the additional which negative I so again shit toe left to minimize the stress produced by additional which negative ions mg which magnesium hydroxide forms until the reaction caution becomes equal to the soul ability product at the new equilibrium, the concentration for MG to positive iron on hydroxide ayan is greater as compared to the initial MD which so and more solid mg, which is present in scenario. Second, moving on to auction. See, In this case, the action shifts toe right toe in Greece. In this case, the action shift to the right to increase the which negative ion concentration as hydroxide start combining with the hydrogen nine to form water molecules. In this case, more mg which is dissolved. So more mg which two is deserved Auction d and me and No. Three does not compete any similar line present in the solution off mg, which hence it has no appreciable effect. So it doesn't effect anything, so it gives us no, I think auction E the additional mg, which has no effect upon the soil ability or on the concentration according to the expression off portion, which is equal to concentration off magnesium too positive on DDE, which make it divine will square. There is no change in concentration, hence changing the amount off Solan. Magnesium hydroxide in the mixture has no effect for the values they're. Therefore, there will be no shifting off reaction.

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