approximately-014quad-mathrmg-of-nickelii-hydroxide-mathrmnimathrmoh_2s-dissolves-per-liter-of-water

Name: approximately 014quad mathrmg of nickelii hydroxide mathrmnimathrmoh_2s dissolves per liter of water
Uploaded: 2020-06-07
Duration: 3 min 18 s
Channel: Numerade
Description: Approximately 0.14$\quad \mathrm{g}$ of nickel(II) hydroxide, $\mathrm{Ni}(\mathrm{OH})_{2}(s),$ dissolves per liter of water at $20^{\circ} \mathrm{C}$ . Calculate $K_{\mathrm{sp}}$ for $\mathrm{Ni}(\mathrm{OH})_{2}(s)$ at this temperature.

Question

Approximately 0.14$\quad \mathrm{g}$ of nickel(II) hydroxide, $\mathrm{Ni}(\mathrm{OH})_{2}(s),$ dissolves per liter of water at $20^{\circ} \mathrm{C}$ . Calculate $K_{\mathrm{sp}}$ for $\mathrm{Ni}(\mathrm{OH})_{2}(s)$ at this temperature.

Anatole Borisov · Accepted Answer

So in this question we&#x27;re looking at, um, the dissociation of nickel hydroxide, um, nickel to hydroxide, to be specific, just solid to become the nickel two plus ion and to hydroxide ions, which are both, uh, acquis. I&#x27;m just having a just traded up here so we don&#x27;t take this now, we&#x27;re going to work through this all symbolically and then just plugging numbers is so we&#x27;re gonna do a nice table to represent the, uh, amount of yeah, that we&#x27;re gonna have a nickel and hydroxide adequate. So we have I see initial change in equilibrium. We ignore solids for this. So doesn&#x27;t matter what&#x27;s over here. Initially, we don&#x27;t have effectively zero concentration of either of these that we&#x27;re gonna have a change, which is gonna be plus acts. And plus two X, which comes from the fact that for every single mole of nickel hydroxide that dissolves, we&#x27;re gonna get a mole of nickel and two moles of hydroxide and that Saudi government of X and two X. So in a different color, we&#x27;re gonna write out the mass actions fresh. Now, remember, the mass action expression is equal to the concentrations of the products here And since we&#x27;re ignoring this, we always go in solids. In the mass action expression, there&#x27;s really nothing in our genome air, nothing for the react. So we&#x27;re gonna have nickel two plus times a concentration of hydroxide squared which comes from the coefficient to here. And so if we plug in this X for nickel and two X for hydroxide will get X times two x squared, which equals four. Execute. So this is we&#x27;re trying to find CASS. Be if we can figure out X that fantastic and were given acts just indirectly works hold different color here. We&#x27;re told that we have 0.14 grams of nickel hydroxide. And so if we could figure out how many moles of nickel hydra because we have that much that dissolves in water Actually, Mom, about that we have 0.14 grams of dissolves in water. So what that means is we can figure out how many moles of that that is. Then we can figure out how many moles of nickel hydroxide dissolved in water, which also is you to be equal to the moors of nickel two plus that&#x27;s dissolved in water, which is equal to acts. And so we need to, uh, divide, uh, 12.14 by the molten black lor weight of nickel hydroxide, which is 92.7 grams per mole. And we&#x27;ll get some amount, which is equal to 1.51 times 10 to negative third moles. And that&#x27;s equal to acts like I just described. And so now, if you plug in that value for four execute, you&#x27;ll get that Casspi is equal to 1.3 seven times 10 to the negative eighth, and that&#x27;s our final.

Aadit Sharma · Answer

First we calculate the solid ability in malls for leader. So the former Fuhrer, for solly ability they will be equal to given weight off. The salute in this case were provided by 0.14 graham off nickel hydroxide divided by one liter off solution multiplied wit. One more off salute that is nickel hydroxide divided by Di Modica Lemus, which is 92 point 71 grams. Using this, the solid ability turns out to be 1.5 multiplied with 10 days to power. Negative three moles. Our leader, now for their own will consider the equilibrium for the reaction and draw an I c E table. Here is the representation off the reaction and I c e table is already provided So inviting the next equation that this Saul ability constant gai sp substituting in the values that this concentration off nickel too positive multiplied with hydroxide concentration we have s for nickel and us square for hydroxide giving us four sq So substituting these values into the individual concentration off 1.5 we have gai sp equals four multiplied with 1.5 10 days to pawn negative three Q So in this case, we obtain a value off 1.4 multiplied 10 days to plan negative eight

David Collins · Answer

to determine Caspi Moeller soluble ity. We&#x27;ve been given the soluble itI in units of grams for leader, but we can convert grams for leader to molds for leader. Using the molar mass of nickel to hydroxide, we get 1.51 times 10 to the negative three moulds for leader has the Moeller Saudi ability because nickel hydroxide separates into one Nikolai on and to hydroxide ions. The concentration of nickel will be equal the Moeller Saudi ability But the concentration of hydroxide will be to time Mohler Celje bility. This then gives us a chaos devalue equal to the nickel concentration which is Mueller soluble ity multiplied by the hydroxide concentration which is two times more soluble Ity square. So now we can plug in our s value of 1.51 times 10 to the negative three and calculate Caspi is 1.4 times 10 to the negative eight

Anatole Borisov · Answer

just the question. I asked us to consider the dissolving of iron too hydroxide in water. And so we have this being an equilibrium with iron two plus Hey, Chris to hydroxide ions. Also a Prius. And so we are given the Moeller solid ability there were given the solid ability in terms of grams per leader of iron hydroxide which we can use. I&#x27;m with blood for weight of iron hydroxide to find the Moeller soluble ity. And then from there we can find the cast. And so we&#x27;re right on a nice table. Quickly. Um, if you&#x27;re not comfortable doing this, go back and do some of the earlier problems that are complicated. And so we start with zero effectively class axe plus two x know that there&#x27;s two hydroxide. So do you Clever enough, accent, You action So Casspi our mass actor expressions giving the concentration of iron two plus times your concentration hydroxide Myers hydroxide squared which, if you plug in, will get acts times two x squared, which equals for excuse. And so if we want to find Caspi which so we&#x27;re asked to do be great if we could figure out what X is and when we actually can. Because we&#x27;re told that we have 1.5 times 10 to the negative there grams per leader of iron hydroxide right here it dissolves. And so if we can figure out the amount of moles that this is over here, that&#x27;s going to be equal to the amount of moles of iron to plus and thus equals X. Explain that again, since we have a 1 to 1 relationship between the amount of iron hydroxide, the dissolves and the amount of iron to plus and one solution. And since the amount of iron two plus in solution, equilibrium is X by figuring out the amount of moles that 1.5 times hence a negative third grams of iron hydroxide this we can figure out X and thus Caspi. So how are we gonna do that? Well, we&#x27;re gonna use the molar masses So 1.5 time sense in negative 30 grams per leaders. If we divide by the molecular weight, which is part of me, this needs to be a division. I do not want to do you guys dirty and it wrong here 89.86 grams per mole the grams cancel out in the value that we get is going to be 1.6 times tense, a negative fifth most for leader, which is also you go to polarity. And so that&#x27;s are equal value of X. Right here you go to X. And so if we&#x27;re looking for Casspi, which is equal to four x cubed well, the problem gets a lot simpler here. Plug in that value. And when you do that, SP is gonna be equal to about two times tense in 1914. There&#x27;s no units on that because equilibrium constants don&#x27;t have yet, so yes, sir. Yes, sir.

David Collins · Answer

This calculation is pretty straightforward if we write Caspian terms of Moeller Celje bility, but let&#x27;s first calculate Mueller scalability. If we know that 1.78 grams dissolved for leader, we can convert the grams into leaders by dividing by the molar Mass and we&#x27;ll get 2.40 times 10 to the negative To Moller for that Mueller Celje bility Caspi is going to be equal to the calcium concentration multiplied by the hydroxide Concentration squared calcium concentrations gonna be equal to the Moeller Celje bility hydroxide. Concentration because we get to hydroxide is for everyone. Calcium hydroxide that dissolves is going to be equal to two times the Moeller Celje bility and then we&#x27;ll square that whole thing. So K SP is going to be equal to 5.55 times 10 to the negative five.

David Collins · Answer

to calculate K sp will recognize that the case P is going to be equal to the calcium concentration multiplied by the hydroxide concentration squared because of the strike geometry. Every time we get a calcium in solution from calcium hydroxide, we get to hydroxide is in solution. Therefore, the calcium concentrations gonna be 1/2 the hydroxide concentration and the hydroxide concentration we can calculate from ph hydroxide concentration will be equal to the hydro knee um concentration divided into K w or kw divided by 10 to the negative pH and we get a hydroxide concentration of 4.79 10 of the negative to. So KSK P is going to be equal to half of that multiplied by the 4.79 times. Tend the negative two squared and we get 5.48 10 to the negative five.

Problem

For the reaction $\mathrm{N}_{2}(g)+3 \mathrm{H}_…

Problem 112 Hard Difficulty

Approximately 0.14$\quad \mathrm{g}$ of nickel(II) hydroxide, $\mathrm{Ni}(\mathrm{OH})_{2}(s),$ dissolves per liter of water at $20^{\circ} \mathrm{C}$ . Calculate $K_{\mathrm{sp}}$ for $\mathrm{Ni}(\mathrm{OH})_{2}(s)$ at this temperature.

Answer

$K_{\mathrm{sp}}=1.35 \times 10^{-8}$

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$K_{\mathrm{sp}}=1.35 \times 10^{-8}$

Introductory Chemistry

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Stephen P.

Allea C.

Karli S.

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Steven S. Zumdahl, Donald J. DeCosteIntroductory Chemistry

Nadia L.

Stephen P.

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Introductory Chemistry