I didn't do very well on this topic in the 1/2 exam so finding revision a little difficult!
Please help with this question, not sure what values to use in what equation
What volume of 0.460 M H2SO4 is required to neutralise 24.00 mL of 0.620 M NaOH?
Thanks in advance !!
So if the H2SO4 is neutralising the NaOH, the n(H+) = n(OH-)
Since you're given both the volume and concentration, you can find out the number of NaOH
n=cV, n=0.620*0.02400 (as volume is in litres)
n=0.01488 = 0.0149 (3 sig figs)
n (OH-) = n (NaOH) = 0.0149
n (H+) = 2*n(H2SO4) as there are two hydrogen ions in each molecule
n(OH-)=n(H+) = 0.0149
n(H2SO4) = 1/2 n(H+) = 0.0744
V(H2SO4) = n/c = 0.0744/0.460 = 0.0162L = 16.2mL
Another route is n(H+) = n(OH-)
c(H+)*V(H+)=c(OH-)*V(OH-)
As the concentration of the H+ ions is twice that of H2SO4,
c(H2SO4)*2*V(H2SO4)=c(NaOH)*V(NaOH)
Then substitute values
So start with the part that you have the most information about, see what you need to do and go from there.
Bit rusty but hope this helps