Titrations

A 250cm cubed solution of NaOH was prepared. 25.0cm cubed of this soltion required 28.2cm cubed of 0.100 mol dm-3 HCl for neutralisation. Calculate what mass of NaOH was dissolved to make up the original 250cm cubed soltion. HCl + NaOH > NaCl +H2O

I've converted the volumes into dm, and then used the 28.2 in dm * 0.1 concentration of HCl to find the moles of HCl. 1 mole of HCl = 1 mole of NaOH. I've used the moles of NaOH * Mr (of NaOH) to get mass, and then times this all by 10 to get to the mass for the 250 solution. I get 5.2, but the answer booklet say 4.1. Where have I gone wrong in my method?

Moles HCl = 0.0282 x 0.1 = 0.00282

this equivalent to 0.00282 moles of NaOH ( in 25cm^/3)

thus 0.00282 x 10 moles NaOH in 250 cm^/3
= 0.0282 moles
RM of NaOH = 40 therefore mass of NaOH = 0.0282 x 40 = 1.128 g / 250 cm^/3

BUT if you made up a litre originally (rather than 250cm^/3) you would need 1.128 x 4 = 4.512g

I disagree with you AND the markbook

Yes, firstly, calculate number of mols of NaOH. Secondly, you know that the acid and the base will combine in a one to one ratio due to the fact that the acid is monoprotic. Thus you will know the number of mols of the acid which have reacted. Then from that you can work out the number of mols of acid in the entire 250ml solution. Then using the number of mols and the mass of the acid, you can work out the relative molecular mass.

87.8?

That's what I get.

Thanks.

A solution of a metal carbonate, M2CO3, was prepared by dissolving 7.46 g of the anhydrous solid in water to give 1000 cm3 of solution. 25.0 cm3 of this solution reacted with 27.0 cm3 of 0.100 mol dm-3 hydrochloric acid. Calculate the relative formula mass of M2CO3 and hence the relative atomic mass of the metal M.

Last question: I get an atomi mass of 4.5 by:

0.027 * 0.1 = moles of HCl
Moles of HCl * 40 = Moles of M2CO3

7.46 / 0.108 = 69
69 - ( 12 + 48) = 9
9/2 = 4.5

Where have I gone wrong?

message too short

moles of HCl = 0.027 x 0.1 = 0.0027

2 moles HCl is equivalent to 1 mole M₂CO₃

In titre moles M₂CO₃ = 0.0027/2

Moles per litre = 0.0027 x 40/2 = 0.054 moles

this is equivalent to 7.46 g

therefore 1 mole = 7.46/0.054 =138

Mass of CO₃ = 60

Therefore remaining mass = 2 x M = 138 - 60 = 78

Therefore M = 39 (Potassium)

Woah. One year later, I am struggling on this question. I made the same mistakes as well... gosh if it wasn't for this I wouldn't have got my A for the homework.

Moles HCl = 0.0282 x 0.1 = 0.00282

this equivalent to 0.00282 moles of NaOH ( in 25cm^/3)

thus 0.00282 x 10 moles NaOH in 250 cm^/3
= 0.0282 moles
RM of NaOH = 40 therefore mass of NaOH = 0.0282 x 40 = 1.128 g / 250 cm^/3

BUT if you made up a litre originally (rather than 250cm^/3) you would need 1.128 x 4 = 4.512g

I disagree with you AND the markbook