H2SO4 + 2KOH ==> K2SO4 + 2H2O
mols H2SO4 initially = M x L = about 0.23 but you need to do it more accurately.
mols KOH added = M x L = about 0.135 and that will neutralize 2*0.135 mols H2SO4 = about 0.27 mol H2SO4. You have less H2SO4 than that; therefore, no H2SO4 will remain.
0.550 L of 0.420 M H2SO4 is mixed with 0.500 L of 0.270 M KOH. What concentration of sulfuric acid remains after neutralization?
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