Solubility chemistry, Molarity and electrolytes

Question 1

What is an electrolyte?

Answer 1

Something that dissociates in a polar solvent such as water to create ions (both cations and anions). Example: Salt dissolves in water and becomes an electrolyte, electrolytes also generate electricity

Question 2

Define solute.

Answer 2

A solute is something that dissolves in a solvent thus creating a solution.

Question 3

Define solvent.

Answer 3

A substance that dissolves a solute creating a solution.

Question 4

Define solution.

Answer 4

A solution is a homogeneous mixture where a solute dissolves in a solvent.

Question 5

Define aqueous solution.

Answer 5

An aqueous solution is a solution where the solvent is water.

Question 6

Define solubility

Answer 6

It is the ability for a solute to dissolve in a solvent and form a solution

Question 7

What goes up when you increase the temperature of a solute and solvent?

Answer 7

The dissolution and solubility increase.

Question 8

What goes up when you increase the surface area of a solute and solvent (crushing solute into powder)?

Answer 8

The dissolution increases.

Question 9

What’s the difference between a strong electrolyte a weak electrolyte and a nonelectrolyte?

Answer 9

A strong electrolyte can conduct electricity very well, strong acids, strong bases and soluble ions ionize completely in water. A weak electrolyte can conduct a little electricity, usually an insoluble compound, weak acid or weak base. A nonelectrolyte cannot conduct electricity, it dissolves in water but never ionizes. .

Question 10

How do you calculate molarity?

Answer 10

Molarity = (mols of solute) / (Litres of solutions)

Question 11

How do you calculate the molarity of a specific element if given the g of compound and L of solution?

Answer 11

Molarity = (G. of compound) / (Molar mass of compound) * (Molar ration between element and compound) / (L. of solution)

Question 12

How do you calculate (in grams) the mass of a substance if you know the litres of solution and molarity?

Answer 12

G = (Molarity) * (L. of solution) * (Molar mass of substance)

The question assumes you know what the substance is.

Question 13

If you want to know the L of solution required to dissolve a known number of grams of a substance and the end molarity, how do you calculate it?

Answer 13

L = (G. of substance) / (Molar mass of substance) / (Molarity)

The question assumes you know what the substance is.

Question 14

If you know the molarity of a compound, how do you calculate the molarity of one of the elements in the compound?

Answer 14

(Molarity of Element) = (Molarity of compound) x (Subscript of that element)

Question 15

What is the formula for molarity dilution?

Answer 15

M1V1 = M2V2
M = Molarity
V = Volume

Question 16

If you know the starting molarity of a substance and the volume of solution to begin with, how do you get a new molarity if you added x numbers of litres of solvent?

Answer 16

M1V1 = M2V2
(Starting Molarity)(Starting Litres) = M2(Starting Litres + X)
M2 = (Starting Molarity) / ((Starting Litres + X) / (Starting Litres))

Question 17

If you’re given grams of a compound and L of solution, how do you calculate the molarity of a specific element?

Answer 17

Molarity = (G. of compound) / (Molar mass of compound) x (Subscript of element) / (Litres of solution)

Question 18

If you’re given the starting molarity, the end molarity and the final volume, how can you find the amount of water to begin with?

Answer 18

M1V1 = M2V2
(Starting Molarity)V1 = (End Molarity)(End Volume)
V1 = ((End M)(End V))/(Start M)

Question 19

When you mix two solutions, know the litres of each solution and the molarity of the solutes, how can we find the molarity of each end solute?

Answer 19

M1V1 = M2V2
Assume X is solution 1 and Y is solution 2
(Molarity of substance in X)(Litres of X) = M2(Litres of X + Litres of Y)
Calculate M2 and repeat for each solute you need to find the molarity of.

Question 20

How do you find the molarity of a solution that is a mix of two other solutions that have the same solute but different molarities and volumes?

Answer 20

Say the first solution is X and the second solution is Y
M1V1 + M2V2 = M3V3
(M1V1 + M2V2)/V3 = M3
((Molarity of X)(L. of X) + (Molarity of Y)(L. of Y)) / (L. of X + L. of Y) = M3

If the volume is the same, it’s the middle of the two volumes.

Question 21

Which elements/ions are always soluble?

Answer 21

Lithium, Sodium, Potassium, Cesium, Rubidium, Nitrate (NO3^-) Ammonium (NH4^+) and Acetate (C2H3O2^-)

Question 22

Which elements are generally soluble, which elements are exceptions?

Answer 22

Cl^+ , Br^+, I^+ are generally soluble unless paired with Ag^+, Pb^2+, Hg2^2+

Sulphate is also generally soluble but it’s exceptions include: Ca^2+, Ba^2+, Sr^2+, Pb^2+

Question 23

What polyatomic ion is generally insoluble, what are the exceptions?

Answer 23

OH^- (Hydroxide)
Exceptions:
Na^+, NH4^+ (Ammonium), Ca^2+, Sr^2+, Ba^2+

Question 24

Which ions are generally insoluble? What are their exceptions?

Answer 24

S^2-, CO3^2-, PO4^2-, CrO4^2-
Exceptions:
Na^+, K^+, Li^+, NH4^+

Question 25

How do you calculate molality?

Answer 25

(Mols of Solute) / (Kg of Solvent)

Question 26

If you’re given the molarity of the first solution and the molarity AND the volume of the second solution, as well as the substance in the solutions, how can we find the volume in the first solutions.

Answer 26

First, write a BALANCED chemical equation of the substance in the first solution reacting with the substance in the second.
Now:
(Molarity of second solution) * (Litres of second solution) * ((Coefficient of first substance) / (Coefficient of second substance)) / (Molarity of first substance)

You can also use (Coefficient of solution 2)M1V1 = (Coefficient of solution 1)M2V2 and find V1

Question 27

Knowing the molarity, litres and substance of the first solution required to react with a volume of the second solution, calculate the molarity of the second solution.

Answer 27

Write a BALANCED chemical equation, the use the formula:
(Coefficient of solution 2) * (Volume of solution 1) * X = (Coefficient of solution 1) * (Volume of solution 2) * (Molarity of solution 2)

Then solve for X
(Solved)X = (Solved)

X can be on both sides. Location depends on the equation, not the wordin

Question 28

How can you calculate the mass of a known product given the volume, molarity and substance that is the first solution and that the second solution is an excess reactant?

Answer 28

First, write a BALANCED chemical equation.
After this, the excess reactant can be discarded. Now we use the following:
(Molarity of solution) * (Litres of solution) * ((Coefficient of product) / (Coefficient of solution))
* (Molar mass of product)

Remember, the excess reactant is discarded after the equation is written

Question 29

How can you calculate the mass of a known product, given the volume, molarity and substance of BOTH solutions.

Answer 29

First, write a BALANCED chemical equation.
Now we use the following for BOTH solutions:
(Molarity of solution) * (Litres of solution) * ((Coefficient of product) / (Coefficient of solution))
* (Molar mass of product)

The answer that is LOWER is the correct answer

The solution is not specified sicne you do it to both solutions.

Question 30

In the equation M1V1 = M2V2, what is the simplest way to use it in acid titration problems?

Answer 30

Set one side to the acid and the other side to the base. It also makes coefficients easier since you don’t need to write a chemical equation, you just balance the two as it is and use the coefficients are the corresponding sides.