Chemistry Question! Answer both or one if you can.?

Question

If 2.0 g of NaOH dissolved in 250 ml of water, what will be the molarity of the resulting solution?

Show me how you did this too please. Thanks!


AND...
If the endpoint in the titration of HCl with NaOH is mistakenly passed (to pink) what effect (high, low, no effect) does this have on the calculated molarity of the NaOH? And explain it to me please.

Thanks everyone!

Answer 1

first things first.. Molarity = Mol / L You know grams and mL. So....

NaOH = 22.99 + 16.00 + 1.08 = 40.00 g/mol (sum weights from periodic table)

2.0g NaOH * 1mol / 40 g = 0.05 mol (1/2 way there)

250mL * 1L/1000mL = 0.25L (convert mL to L)

0.05 g / 0.25 L = 0.2M

For the titration question, it sounds like you're blowing by the endpoint (color change) in doing the titration. You need to understand that you're trying to find out how strong the HCl acid is by diluting it with NaOH base. If you've added too much base, you would mistakingly think that the acid is stronger than it really is. So you would be effectively saying that it has a HIGHER molarity than actual. Explained mathematically as above by molarity = mol / L If you think the acid is stronger, you've got MORE mol / L or more grams.

Answer 2

The molarity is the concentration, in moles per unit volume. Moles per litre (cubic decimetre) is the common measure.
So, How much does a mole of NaOH weigh?
The molar mass of sodium: 23 g/mol
O: 16 g/mol
H: 1 g/mol
So, the molar mass of NaOH = 40 g/mol. ie, 1 mole of NaOH weighs 40 g.
so, how many moles in 2.0 g of NaOH?
2.0 / 40 = 0.05 moles.
You have 0.05 moles of NaOH in 250 ml of water.
How much does 1 ml of this solution contain?
1/250 of 0.05.
0.05/250 = 0.0002 moles per ml
How much is this per litre?
1000 ml = 1L. so, 0.0002 x 1000 = 0.2 mol/L

If you titrate HCL with NaOH, then, at the end point, the NaOH concentration falls to zero. If you pass the end-point, you put in MORE HCl than is needed to neutralise the solution. If you are using the amount of HCl used to calculate the amount of NaOH present, then you will end up with a higher than accurate value for the NaOH concentration.