Solubility: 1. The volume of water needed to dissolve 0.0589 grams of calcium ch

Question

Solubility:

1. The volume of water needed to dissolve 0.0589 grams of calcium chromate is ____ L.

(Assume no volume change upon addition of the solid.)

2. The mass of magnesium hydroxide that is dissolved in 225 mL of a saturated solution is ______ grams.

3. Consider these compounds:

Complete the following statements by entering the letter corresponding to the correct compound.

Without doing any calculations it is possible to determine that iron(II) carbonate is more soluble than __, and iron(II) carbonate is less soluble than __.

It is not possible to determine whether iron(II) carbonate is more or less soluble than __ by simply comparing Ksp values.

4.

Complete the following statements by entering the letter corresponding to the correct compound.

Without doing any calculations it is possible to determine that barium phosphate is more soluble than __, and barium phosphate is less soluble than __.

It is not possible to determine whether barium phosphate is more or less soluble than __ by simply comparing Ksp values.

5. For each of the salts on the left, match the salts on the right that can be compared directly, using Ksp values, to estimate solubilities.

(If more than one salt on the right can be directly compared, include all the relevant salts by writing your answer as a string of characters without punctuation, e.g, ABC.)

Write the expression for K in terms of the solubility, s, for each salt, when dissolved in water.

lead sulfate

lead phosphate

Ksp = ____

Ksp = ____

Note: Multiply out any number and put it first in the Ksp expression. Combine all exponents for s.

Explanation / Answer

Question .1

Dear the some information is missing in the question, there is no information to relate 0.589 gm with solubility. I am putting two general methods here, hope it will solve your problem.

We can measure amount of water if we have solubility product (Ksp) or solubility of Calcium chromate in water. Let us do this by both ways:

Method 1.

Equation is

Cacro4 <------>Ca2+ + Cro42- (saturated solution)

Solubility product (Ksp)

Ksp = [Ca2+][Cro42-]

Ksp=x*x

x2= root Ksp=A (suppose)

Since the equation above shows a 1:1 mole ratio of calcium chromate to Ca+2 ions, we can assume that A moles of Cacro4 will dissolve. Since CacrO4 has a mass of 156.072 g/mol, this would equate to:

(A mol/L) x (156.04 g/mol) =B grams/L

that means B grams of Cacro4 will get dissolve in in 1 liter

Therefore 0.0589 gm will dissolve in =(1 litre/B)* 0.0589gm= C litre

Method 2.

if SOLUBILITY OF Calcium chromate in water is given.

(value given in one of online source is =2.25 g/100 mL (20 °C)

Therefore it means 2.25 grams of compound is dissolved in 100 mL

Therefore for 0.0589 gm will be soluble in (100/2.25)*.0589 =2.61 ml=2.61/1000 ml=0.00261 L

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