25 cm^3 of 2mol/dm^3 Sodium hydrixude solution and 25cm^3 of 2 mol/dm^3 Sulphuric acid were mixed well to make a solution X. Calculate the concentration of Sulphuric acid in the final solution..
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Do you mean sodium hydroxide for ''Sodium hydrixude''?
Can you write the chemical equation for the reaction first?
2NaOH + H2SO4 ---> Na2SO4 + 2H20
In this reaction, what is the limiting agent?
limiting agent means? i found the answer it is 1 moldm^-3 cn u check it
Hmm, I didn't get the same answer as you.
can u shw ur working
Perhaps I didn't do it correctly, since I haven't studied chemistry for quite a long time..
No. of mole of H2SO4 used = 1/2 (no. of mole of NaOH) = (1/2) (2x25/1000) = x
*NaOH is the limiting agent that will be used up in the reaction.
No. of mole of H2SO4 left = no. of mole of H2SO4 - no. of mole of H2SO4 used
= (2x25/1000) - x
Concentration of H2SO4 in the final solution = no. of mole of H2SO4 left / volume of the solution.
And I take volume of the solution remains unchange, that is 25cm^3 + 25cm^3 = 50cm^3
Did I do anything wrong?
This is how i did, check if im wrong
2NaOH + H2SO4 ---> Na2So4 +2 H20
the molar ratio is tht, 2 moles of NAOH reacts with 1 mole of H2so4
when u calculate the number of moles in each reactnt it is 0.05 mole, which means 0.05 mol of Naoh react with 0.025 mol of H2so4...so there is 0.025 mole of H2so4 left
n = c*v/ 1000
0.025 = c * 25/1000
c = 1 mol dm^-3
Why is it 25cm^3 for your volume?
should it be 50 or 25?
im not sure of this
Hmm.. final solution, it's not just the volume of H2SO4, so I prefer using the total volume. But I'm not sure though. As I've said, I haven't studied chemistry for quite a long time!
then the answer becomes 0.5 moldm^-3
That's what I got. Sorry for the late reply
so my method is right?
The method is right, just the volume is not right/