Answer:
0.197 M
Explanation:
The reaction equation is:
H2SO4(aq) +2KOH(aq) ----> K2SO4(aq) + 2H2O(l)
number of moles of H2SO4 = 0.25 L * 0.45 M = 0.1125 moles
number of moles of KOH = 0.2 L * 0.24 M = 0.048 moles
since H2SO4 is the reactant in excess;
2 moles of KOH reacts with 1 mole of H2SO4
0.048 moles of KOH reacts with 0.048 * 1/2 = 0.024 moles of H2SO4
Amount of excess H2SO4 left unreacted = 0.1125 - 0.024 = 0.0885 moles
Total volume = 0.25 L+ 0.2 L = 0.45 L
concentration of H2SO4 = 0.0885/0.45 = 0.197 M
Answer:
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For a neutral solution,
[H+][OH-] = 1 x 10⁻¹⁴
The chemical formula of oxalic is 
When oxalic acid reacts with water, first, oxalic acid removes one proton and results in the formation of mono acids.
After that, in second step, oxalic acid in aqueous solution removes another proton which shows it is a polyprotic acid.
The chemical equations are: (the reactions occurs in two steps due to presence to hydrogen atoms).
When one proton is removed:

When another proton is removed:

The dissociation of oxalic acid in water in shown in the image.