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Determining normality, grams, and missing amounts

Submitted by pomegranate on Tue, 04/28/2009 - 23:43

I've been gone from school for a few days and I could really use some help figuring out how to do these problems-

1) Determine the grams necessary to prepare 100.0 mL of 0.50-N sodium hydroxide.

2) Determine the normality of 3.0 grams of calcium acetate in 200 mL of water.

3) Determine the missing amount in 15.0 mL of 0.20-M sodium hydroxide titrates 10.5 mL of __-M hydrochloric acid.

There are also a few problems like the third one where I have to determine the normality instead of molality, but I'm not sure if that makes a difference.

I'm completely lost, please help!  :-\


1. Find the formula mass of NaOH by plugging atomic masses into the formula - an answer close to 40 should be found (23+16+1 for Na,O and H respectively).
1 mole of NaOH will then weigh this mass expressed in grams - 1 mole weighs 40g. So 1 1M solution will need to have 40g dissolved and made up to 1 litre with pure water. This is the solution expressed as molarity (concentration in moles /L).
However, normality is different and refers to the number of reacting equivalents in the solution. In NaOH, OH is the reacting equivalent. A 1.0M solution of NaOH contains 1 mole NaOH per litre and 1 reacting equivalent of OH per litre - so this is also a 1N solution. So 40g NaOH will make a 1.0N solution if made up to 1 litre. The equivalent mass of NaOH = molar mass/1
A 1.0M solution of Ca(OH)2 contains 1 mole per litre but has 2 reacting equivalents of OH per litre so is 2N (2 Normal). The equivalent mass Ca(OH)2 = molar mass/2 - so about 74/2 = 37g is the mass needed to make 1L of 1N solution
1.0M HCl = 1.0N HCl, 1.0M H2SO4 = 2N H2SO4. The equivalent mass of H2SO4 = molar mass/2 (so 98/2 =49g will be the mass of H2SO4 needed to make 1L of 1N solution.)
This should help with molarity and normality
In the first example, the number of equivalents = v in L x normality = 0.100 x 0.5 = 0.050.
So you will need 0.050 x 40 = 2g NaOH
2. Ca(CH3COO) - the equivalent mass will be the molar mass/2.
Find molar mass (about 99g). Find equivalent mass (about 49.5g). Calculate the fraction of this mass that 3.0 g will be. This will be the number of equivalents in 200 mL. You then just need to number of equivalents in 1 litre (5 times this volume)
3. Write a balanced equation and determine the ratio of moles NaOH and HCl reacting. From the vol and molarity of NaOH in question, find the number of moles (no of moles = v in L x molarity). Relate this to the balanced chemical equation to find the number of moles HCl that are needed to react with this. You now know the number of moles HCl and its volume, so you can find the molarity.