Chemistry help please urgent...measuring solubility

Question

A solution of sodium carbonate has a concentration of 5.2 g L-1. A volume of 120 mL of this solution is diluted with water to 3.0 L. How do i find out the concentration of the new solution?

What volume of 10 M hydrochloric acid would be required to prepare 250 mL of 0.30 M HCl?

How much water must be added to 25 mL of a 4.0 M solution of a potassium carbonate solution to dilute it to 1.5 M?

i know all this is alot but i'm sorry. i urgently need this and any help would be great.

Answer 1

In any problem where you are simply diluting you can use the dilution formula:   C1 x V1 = C2 x V2

C's are any concentration unit (as long as they are the same)

V's are any volume (as long as they are the same)

In your first problem :

C1 = 5.2g/L and V1 = 120 mL

C2 = unknown (x) and V2 = 3.0L

Make sure you either change 120 ml to Liters or 3.0 Liters to mL!

Substitute and solve.

You should be able to do the rest of them; just watch your units.

Answer 2

These are all simply conc reductions...easy to follow if you use the equivalence formula V1C1 = V2C2

1) 120 x 5.2 = 3000 x C...C = 0.208 g/L

2)  V x 10 = 250 x 0.30...V = 7.5 mL

3)  25 x 4.0 = V x 1.5...so FINAL volume must be 67 mL...so ADD 42 mL of water.

Answer 3

Sounds like you're in a jam, Lady Marmalade! However, the problems are only simple arithmetic. The key is that the amount of reactive species does not change. All that changes is the amount of water so that

Volume(1) x Concentration(1) = Volume(2) x Concentration(2)

120 ml of Na2CO3 @ 5.2 gL^-1 contains 120*5.2/1000 = 0.624 g. In 3.0L, this be a concentration of 0.624/3 = 0.208gL^-1

10 x V(1) = 0.3 x 250; V(1) = 0.3 x 25 = 7.5 ml of the 10M HCl. If you want to think it through, note that 250 ml of 0.3M HCl contains 0.075 moles HCl. 1 ml of 10M HCl contains 0.01 moles, so you need 7.5 ml. Most people find it easier to use the formula

V(2) = 25 x 4/1.5 =  66.67 ml, Add 66.67-25 = 51.67 ml H2O