H2SO4+2KOH⟶K2SO4+2H2O Suppose 0.350 L of 0.470 M H2SO4 is mixed with 0.300 L of 0.240 M KOH . What concentration of sulfuric acid remains after neutralization?

Answer 1

H2SO4+2KOH⟶K2SO4+2H2O

So we can see from the equation that two KOH molecules are required to neutralize one H2SO4 molecule. Thus, if we know how many moles of KOH we have, then we can determine how many moles of H2SO4 can be neutralized. We have 0.300L of 0.240 M KOH. 0.3L x 0.240 mol/L = 0.072 mol KOH. If two molecules of KOH are required to neutralize one molecule of H2SO4, then 0.072 mol KOH x (1 H2SO4 / 2 KOH) = 0.036 mol of H2SO4 will be neutralized.

Okay, now we need to know how much H2SO4 we're starting with. From the question, 0.350L x 0.470 mol/L = 0.1645 mol H2SO4.

Total - neutralized = remaining

0.1645 mol - 0.036 mol = 0.1285 mol H2SO4

So 0.1285 mol H2SO4 is dissolved in what volume of solution?

0.350L+0.300L = 0.650L

0.1285 mol H2SO4 / 0.650 L = 0.198 M