Question

Suppose 0.250 L0.250 L of 0.470 M H2SO40.470 M H2SO4 is mixed with 0.200 L0.200 L of 0.220 M KOH0.220 M KOH. What concentration of sulfuric acid remains after neutralization?

Answer 1

volume of H2SO4, V = 0.25 L

use:

number of mol in H2SO4,

n = Molarity * Volume

= 0.47*0.25

= 0.1175 mol

volume of KOH, V = 0.2 L

use:

number of mol in KOH,

n = Molarity * Volume

= 0.22*0.2

= 4.4*10^-2 mol

Balanced chemical equation is:

H2SO4 + 2 KOH ---> K2SO4 + 2 H2O

1 mol of H2SO4 reacts with 2 mol of KOH

for 0.1175 mol of H2SO4, 0.235 mol of KOH is required

But we have 4.4*10^-2 mol of KOH

so, KOH is limiting reagent

According to balanced equation

mol of H2SO4 reacted = (1/2)* moles of KOH

= (1/2)*4.4*10^-2

= 2.2*10^-2 mol

mol of H2SO4 remaining = mol initially present - mol reacted

mol of H2SO4 remaining = 0.1175 - 2.2*10^-2

mol of H2SO4 remaining = 9.55*10^-2 mol

Total volume = 0.250 L + 0.200 L = 0.450 L

Use:

[H2SO4] remaining = mol of H2SO4 remaining / Total volume

= ( 9.55*10^-2 mol) / (0.450 L)

= 0.212 M

Answer: 0.212 M