Question

Part 1: Write the balanced neutralization reaction between H2SO4 and KOH in aqueous solution. Phases are optional.

Part 2: 0.550 L of 0.430 M H2SO4 is mixed with 0.500 L of 0.260 M KOH. What concentration of sulfuric acid remains after neutralization?

Answer 1

Part 1: Write the balanced neutralization reaction between H2SO4 and KOH in aqueous solution. Phases are optional.

H2SO4 + KOH --> H2O + K2SO4

the balnaced equation:

H2SO4 + 2KOH --> 2H2O + K2SO4

2)

V = 0.55 L

M = 0.43 of H2SO4

mol = M*V = 0.55*0.43 = 0.2365 mol of H2SO4

V = 0.5 L

M = 0.26 KOH

mol = 0.5*026 = 0.13 mol of KOH

find excess of sulfuric acid

acid - base/2 = 0.2365 - 0.13/2 = 0.1715 mol of acid are left

M = mol/V = 0.1715 / (0.55+0.5)= 0.1633 M of H2SO4 left