calculations (1 Viewer)

[xeriphic]

could anyone help me with this question

a dam contains 3 x 10^8 literes of H2O the H+ concentration is 2 x 10^-3 molL-1, if the acid is HCl,what mass of NaOH would you need to neutralise?

 

[Paroissien]

Sorry, can't be bothered to do actual calculations
HCl + NaOH ---> NaCl + H2O
Firstly work out how many moles of HCl you have: moles = M=molarity x L
As you can see 1 mole of HCl = 1 mole of NaOH
Now you have the moles of NaOH
mass = moles x M (M = 40)
And there is your answer

 

[tennille]

HCl + NaOH ---> H20 + NaCl

Ratio of HCl to NaOH is 1 : 1

m = n/ v

2 * 10-3 = n/ 3*108

n(HCl) = 600000

therefore, n of NaOH is 600000

mass = molar mass * moles

mass = (22.99 + 1.008 + 16) * 600000

mass = 2.4 * 107 litres

That should be the answer, but you might want to check.

 

[Paroissien]

That is the same answer I would've got. I'm 99% sure that's right

 

[tennille]

kewl...i actually got something right for once.

 

[xeriphic]

lol okai thanks both

though I was thinking about the fact HCl was diluted in water, and should find the original concentration of the HCl before applying the other processes

 

[Paroissien]

You are given the concentration of the HCl (or the H+ anyway) in the dam, although you have probably picked that up.
Did you get the right answer btw?

 

[bobbie212]

tennile, all u did was use the workin from the previous post

 

[tennille]

Did you realise that there was a 2 minute difference in the posting times...do you really believe I could right that in 2 minutes? What do you care anyway...? So before you start throwing accusations at people, think!!!!!

 

[bobbie212]

oh ok my bad

 

[tennille]

sorry...I hate when that happens...it takes a long time to type that stuff up

 

[Paroissien]

Yeah, took me a far while as well. Thankfully I got in first though!

 

[tennille]

Hahaha...