Question

Hi super smart people :D I really need help with this problem, I was sick and missed my Chemistry class yesterday so your help would be highly appreciated! If it takes 50mL of .5 M KOH to completely neutralize 125mL of sulfuric acid solution (H2SO4) what is the concentration of the H2SO4 solution?

Answer 1

Sense sulfuric acid is a strong inorganic acid we may assume it fully dissociate and we can write the reaction:
2 KOH + H2SO4 -> K2SO4 + 2 H2O

From the reaction can we conclude:
\[n(H _{2}SO _{4})=\frac{ n(KOH) }{ 2 }\]
We can see from the information in the assignment we are able to calculate n(KOH):
\[C(KOH)=\frac{ n(KOH) }{ V(KOH) } \rightarrow n(KOH)=C(KOH)*V(KOH)\]
In addtion to that we know that:
\[C(H _{2}SO _{4})=\frac{ n(H _{2}SO _{4}) }{V(H _{2}SO _{4}) }\]

Now you should be able to calculate each part or you can substitute so we get an expression of known variables that would be:
\[C(H _{2}SO _{4})=\frac{ \frac{ C(KOH)*V(KOH) }{ 2 } }{ V(H _{2}SO _{4}) }\]

Answer 2

OMG THANK YOU :D @Frostbite you are awesome ^-^