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Concentration Calculation.. Moles!

Hi guys, could anyone help me with the following question?

A student dissolved 5g of Ca(OH)2 in 250ml of H2O. They then titrated 25cm3 of the Ca(OH)2 with HCl. 33.2cm3 of the HCl was required to neutralise the Ca(OH)2. What is the concentration of HCl?

Reply 1

BJack

What's the balanced neutralization reaction?
How many moles of Ca(OH)₂ were used in the titration?
So how many moles were there in the 33.2 ml of HCl – and therefore what's the concentration?

Reply 2

harryheych

Original post by BJack

What's the balanced neutralization reaction?
How many moles of Ca(OH)₂ were used in the titration?
So how many moles were there in the 33.2 ml of HCl – and therefore what's the concentration?

I believe the balanced neutralization reaction is
Ca(OH)₂+ 2HCl -> CaCl₂ + 2H2O

I think that 0.0067 moles were used and so would the concentration be 0.20 moldm^-3... ?

Reply 3

Dylann

Original post by harryheych

I believe the balanced neutralization reaction is
Ca(OH)₂+ 2HCl -> CaCl₂ + 2H2O

I think that 0.0067 moles were used and so would the concentration be 0.20 moldm^-3... ?

0.0067 moles of Ca(OH)2 were used yes, but look at the equation. Twice as many HCl moles react! So the moles of HCl must be molCa(OH)2 x 2

then you can work out concentration

Reply 4

harryheych

Original post by Dylann

0.0067 moles of Ca(OH)2 were used yes, but look at the equation. Twice as many HCl moles react! So the moles of HCl must be molCa(OH)2 x 2

then you can work out concentration

Oh! Thank you! So, I got 0.0134 moles and found the concentration of HCl to be 0.40 mol dm-3 :smile:

(edited 9 years ago)

Reply 5

Borek

I got 0.0135 and 0.41 - could be we used slightly different molar masses.

I don't get how the student dissolved 5 g of Ca(OH)₂ in 250 mL of water. At the time I studied calcium hydroxide solubility was so low we could dissolve no more than about 0.5 g in such a volume :wink: