Question

Write the balanced neutralization reaction that occurs between H 2 SO 4 and KOH in aqueous solution. Phases are optional. neutralization reaction: H_{2}SO_{4}(aq) +KOH(aq)->K_{2}SO_{4}(aq) +H_{2}O(l) H 2 SO 4 ( aq ) + KOH ( aq ) ⟶ K 2 SO 4 ( aq ) + H 2 O ( l ) Suppose 0.950 L of 0.400 M H 2 SO 4 is mixed with 0.900 L of 0.260 M KOH . What concentration of sulfuric acid remains after neutralization?

Answer 1

Concentration H2SO4 remaining after neutralization = 0.142 M

Step-by-step explanation:

Step 1: Data given

Volume of a 0.400 M H2SO4 = 0.950 L

Volume of a 0.260 M KOH = 0.900 L

Step 2: The balanced equation

H2SO4(aq) +2KOH(aq) → K2SO4(aq) + 2H2O(l)

Step 3: Calculate moles

Moles = molarity * volume

Moles H2SO4 = 0.400 M * 0.950 L

Moles H2SO4 = 0.380 moles

Moles KOH = 0.260 M * 0.900 L

Moles KOH = 0.234 moles

Step 4: Calculate limiting reactant

For 1 mol H2SO4 we need 2 moles KOH to produce 1 mol K2SO4 and 2 moles H2O

KOH is the limiting reactant. It will completely be consumed (0.234 moles). H2SO4 is in excess. There will react 0.234/2 =0.117 moles

There will remain 0.380 -0.117 = 0.263 moles H2SO4

Step 5: Calculate concentration of sulfuric acid remains after neutralization

Concentration H2SO4 = moles H2SO4 / volume

Concentration H2SO4 = 0.263 moles / (0.950L+ 0.900L)

Concentration H2SO4 = 0.142 M