Making a Solution

Question

Make 50.00mL of a solution that is 0.20M KIO₃ and 0.080M H₂SO₄.

How many grams of KIO₃ are needed?

How many mL of 2.0M H₂SO₄ is needed?

Answer 1

a) How many grams of KIO₃ are needed?

First determine the molar mass of KIO₃ (214.001 g/mol)

(50.00mL) x ( 1 L ) x (0.20 mol) x (214.001g) = 2.1 g KIO₃

(1000mL) ( 1 L ) ( 1 mol )

b) How many mL of 2.0M H₂SO₄ is needed?

C₁V₁ = C₂V₂

(50.00 mL) x (0.080 M) = (C₂) x (2.0 M)

Solve for C₂

C₂ = 2.0 mL of 2.0M H₂SO₄ is needed

Answer 2

What you know:

You know that the volume (V) of the solution is 50.00 mL, let's convert this to L to simplify the process.

V = 50.00 mL * (1L/1000 mL) = 0.0500 L

You may also know that molarity (M) represents the concentration of a certain solute within a solution based on the formula:

M = mols of solute / Volume of solution (L)

Since we know the volume of the solution and the Molarities of both compounds we can do some math here to determine the mols (amount) of each compound within the solution.

KIO3: (0.20M)(0.0500 L) = 0.0100 mols KIO3

H2SO4: (0.080 M)(0.0500 L) = 0.004 mols H2SO4

Since we know how many mols of KIO3 are in the solution we can convert using the molar mass (Molecular weight) of KIO3 to get grams of KIO3 needed.

For this we refer to the periodic table and add up the molecular weights of one K, one I, and 3 O: MW (KIO3) = 214 g/mol

So doing some math:

0.0100 mols * (214 g/mol) = 2.14 g KIO3 *****

For H2SO4 we can follow a similar path:

We once again refer to the molarity equation: M = mols of solute/ Volume of solution

Since we know the original molarity of a solution of pure H2SO4 we can use the fact that we know how many mols of H2SO4 end up in the final solution and the original molarity to determine the Volume of pure H2SO4 required.

(0.004 mols H2SO4) / 2.0 M = 0.002 L H2SO4 of 2.0 M required !

Now convert L to mL to get the final answer:

0.002 L * (1000 mL/L) = 2.00 mL of 2.0M H2SO4 required ****

Another way you can get this answer is by referring to the formula: M1V1 = M2V2

M1 refers to the molarity of H2SO4 orginally, V1 refers to the Volume of H2SO4 originally ;

M2 refers to the molarity of H2SO4 final, V2 refers to the Volume of H2SO4 final

Since the number of mols of H2SO4 DON'T change this formula works since Volume of the solution is now the only variable that can change the molarity.

V1 = (M2V2)/M1 = ((0.080 M)(50.00 mL))/(2.0 M) = 2.00 mL of 2.0M H2SO4 *****

Note that the units of Volume in this case doesn't have to be in L since they cancel each other out anyway.