0.550 L of 0.470 M H2SO4 is mixed with 0.500 L of 0.220 M KOH. What concentratio

Question

          0.550 L of 0.470 M H2SO4 is mixed with 0.500 L of 0.220 M KOH. What concentration of sulfuric acid remains after neutralization?           0.550 L of 0.470 M H2SO4 is mixed with 0.500 L of 0.220 M KOH. What concentration of sulfuric acid remains after neutralization?      0.550 L of 0.470 M H2SO4 is mixed with 0.500 L of 0.220 M KOH. What concentration of sulfuric acid remains after neutralization? 0.550 L of 0.470 M H2SO4 is mixed with 0.500 L of 0.220 M KOH. What concentration of sulfuric acid remains after neutralization?

Explanation / Answer

H2SO4 + 2 KOH => K2SO4 + 2 H2O


Initial moles of H2SO4 = 0.550 x 0.470 = 0.2585 mol

Initial moles of KOH = 0.500 x 0.220 = 0.11 mol


Moles of H2SO4 reacted = 1/2 x moles of KOH

= 1/2 x 0.11 = 0.055 mol


Excess moles of H2SO4 = 0.2585 - 0.055 = 0.2035 mol

Total volume = 0.550 + 0.500 = 1.05 L


Concentration of H2SO4 remaining = excess moles of H2SO4/total volume

= 0.2035/1.05

= 0.194 M

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