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The balanced chemical equation between h2so4 and koh in aqueous solution ; H2SO4 (aq) + 2KOH (aq) --------> K2SO4 (aq)+ 2H2O (l)
Next, calculate moles of each reactant:
0.650 L X 0.430 M H2SO4 = 0.2795 mol H2SO4
0.600 L X 0.240 M KOH = 0.144 mol KOH
Next, calculate moles of H2SO4 consumed by the KOH:
1 eq of H2SO4 reacts with 2 eq of KOH
0.144 / 2 M KOH = 0.072 M of H2SO4 would be reacted
So, remaining moles of H2SO4 = 0.2795 - 0.072 = 0.2075 mol H2SO4
This is in a total volume of 1.25 L, so the concentration of H2SO4 is:
0.2075 mol / 1.25 L = 0.166 M
As a base, KOH is very reactive, while H2SO4 is a highly reactive acid. If a dilute solution of each acid and base is combined, a neutralization reaction will occur.
In this example, the neutralization process would produce water as a byproduct and K2SO4 as the salt as its end products.
Learn more about to chemical equation visit here;
https://brainly.com/question/28294176
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