Write the balanced neutralization reaction between H2S04 and KOH in aqueous solution. Phases are optional. 0.250 L of 0.470 M H2S04 is mixed with 0.200 L of 0.240 M KOH. What concentration of sulfuric acid remains after neutralization? Number M H,so, Solution 1. H2SO4 + 2KOH = K2SO4 + H2O 2.1 mole of H2SO4 reacts with 2 mole of KOH. So molarity of H2SO4 requires Now Moles of H2SO4 = molarity * volume = 0.250 * 0.470 = 0.1175 moles Moles of KOH = 0.200*0.240 = 0.048 moles So mole of H2SO4 requires = 0.048 / 2 = 0.024 moles Remaining moles of H2SO4 = 0.1175 - 0.024 = 0.0935 moles Concentration = moles / total volume = 0.0935 / ( 0.250 + 0.200 ) = 0.21 M If you have any query please comment If you satisfied with the solution please rate it thanks .

1. Write the balanced neutralization reaction between H2S04 and KOH in aqueous solution. Phases
are optional. 0.250 L of 0.470 M H2S04 is mixed with 0.200 L of 0.240 M KOH. What
concentration of sulfuric acid remains after neutralization? Number M H,so,
Solution
1. H2SO4 + 2KOH = K2SO4 + H2O
2.1 mole of H2SO4 reacts with 2 mole of KOH. So molarity of H2SO4 requires
Now
Moles of H2SO4 = molarity * volume
= 0.250 * 0.470
= 0.1175 moles
Moles of KOH = 0.200*0.240 = 0.048 moles
So mole of H2SO4 requires = 0.048 / 2 = 0.024 moles
Remaining moles of H2SO4 = 0.1175 - 0.024
= 0.0935 moles
Concentration = moles / total volume
= 0.0935 / ( 0.250 + 0.200 )
= 0.21 M
If you have any query please comment
If you satisfied with the solution please rate it thanks