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Leya [2.2K]
3 years ago
5

Calculate the molar solubility of CaF2 at 25°C in a solution that is 0.010 M in Ca(NO3)2. The Ksp for CaF2 is 3.9 x 10-11.

Chemistry
1 answer:
madreJ [45]3 years ago
4 0

Answer:

Molar \ solubility=3.12x10^{-5}M

Explanation:

Hello,

In this case, for the dissociation of calcium fluoride:

CaF_2(s)\rightleftharpoons Ca^{2+}+2F^-

The equilibrium expression is:

Ksp=[Ca^{2+}][F^-]^2

In such a way, via the ICE procedure, including an initial concentration of calcium of 0.01 M (due to the calcium nitrate solution), the reaction extent x is computed as follows:

3.9x10^{-11}=(0.01+x)(2*x)^2\\\\x=0.0000312M

Thus, the molar solubility equals the reaction extent x, therefore:

Molar \ solubility=3.12x10^{-5}M

Regards.

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Express the concentration of a 0.0320 M aqueous solution of fluoride, F−, in mass percentage and in parts per million (ppm). Ass
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Answer:

607 ppm

Explanation:

In this case we can start with the <u>ppm formula</u>:

ppm=\frac{mg~of~solute}{Litters~of~solution}

If we have a solution of <u>0.0320 M</u>, we can say that in 1 L we have 0.032 mol of F^-, because the molarity formula is:

M=\frac{mol}{L}

In other words:

0.0320~M=\frac{mol}{1~L}

mol=0.032~M*1~L=0.032~mol

1~L~of~Solution=0.0320~mol~of~solute

If we use the <u>atomic mass</u> of F  (19 g/mol) we can convert from mol to g:

0.0320~mol~F^-\frac{19~g~F^-}{1~mol~F^-}~=~0.607~g

Now we can <u>convert from g to mg</u> (1 g= 1000 mg), so:

0.607~g\frac{1000~mg}{1~g}=607~mg

Finally we can <u>divide by 1 L</u> to find the ppm:

ppm=\frac{607~mg}{1~L}=~607~ppm

<u>We will have a concentration of 607 ppm.</u>

I hope it helps!

4 0
3 years ago
As the temperature of one of the four liquids increases, the vapor pressure will
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Answer:

a) increase exponentially.

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The vapor pressure is depend only on temperature.

The vapor pressure of liquid does not depend upon amount of liquid. For example whether the liquid is 50 g or 30 g its vapor pressure will remain same according to the temperature.

The temperature and vapor pressure have exponential relationship. As the temperature of liquid increases its vapor pressure also goes to increase. When the temperature of liquid goes to decrease its vapor pressure also decreases.

The change in vapor pressure of substance when temperature changes is given as,

ln P₂/P₁ = ΔH(va)/R (1/T₁ - 1/T₂)

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