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kondor19780726 [428]
3 years ago
8

Using your value of Ksp, and starting with an equilibrium system consisting of a saturated solution of calcium hydroxide, predic

t whether the solubility will increase or decrease it you add a. A. 5.00 mL of 0.100 M HNO3 Ksp = 2.966610-5 b. 5.00 mL of 0.100 M NaOH C. 5.00 mL of 0.100 M Ca(NO3)2
Chemistry
1 answer:
Ugo [173]3 years ago
6 0

Answer:

(A) HNO_{3} Solubility will increase.

(B) NaOH Solubility will decrease.

(C) Ca(NO_{3} )_{2} Solubility will decrease.

Explanation:

a) According to Le Chatelier's Principle when a change is introduced in a reaction system at equilibrium, the system responds by the reaction shifting in the direction to counter the change introduced.

When the change introduced is addition of acid, some OH^{-} are consumed due to it and the system would respond by reaction shifting towards creation of more OH^{-} in the system by increased dissolution. Thus solubility increase.

(b) When OH^{-} are added, the reaction shifts to counter the change introduced of increased OH^{-} by shifting to side that decreases the increased OH^{-}. That happens by reaction shifting towards some Ca (OH)_{2} precipitating back. Hence solubility decreases.

(c) The change introduced is addition of Ca^{2+} which will be countered by reaction shifting to side that decreases increased Ca ^{2+}, which happens by some Ca(OH)_{2} precipitating back. Hence solubility decreases.

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Answer : The percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

Explanation :

First we have to calculate the number of atoms in 5.5 wt% Pb and 94.5 wt% of Sn.

As, 207.2 g of lead contains 6.022\times 10^{23} atoms

So, 5.5 g of lead contains \frac{5.5}{207.2}\times 6.022\times 10^{23}=1.59\times 10^{22} atoms

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As, 118.71 g of lead contains 6.022\times 10^{23} atoms

So, 94.5 g of lead contains \frac{94.5}{118.71}\times 6.022\times 10^{23}=4.79\times 10^{23} atoms

Now we have to calculate the percent composition of Pb and Sn in atom.

\% \text{Composition of Pb}=\frac{\text{Atoms of Pb}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100

\% \text{Composition of Pb}=\frac{1.59\times 10^{22}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=3.21\%

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\% \text{Composition of Sn}=\frac{\text{Atoms of Sn}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100

\% \text{Composition of Sn}=\frac{4.79\times 10^{23}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=96.8\%

Thus, the percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

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