Question

Write the balanced neutralization reaction that occurs between H2SO4 and KOH in aqueous solution. Phases are optional. 

neutralization reaction: 

Suppose 0.750 L of 0.400 M H2SO4 is mixed with 0.700 L of 0.280 M KOH. What concentration of sulfuric acid remains after neutralization? 

Answer 1

1)

Balanced chemical equation is:

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

2)

volume of H2SO4, V = 0.75 L

use:

number of mol in H2SO4,

n = Molarity * Volume

= 0.4*0.75

= 0.3 mol

volume of KOH, V = 0.7 L

use:

number of mol in KOH,

n = Molarity * Volume

= 0.28*0.7

= 0.196 mol

Balanced chemical equation is:

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

1 mol of H2SO4 reacts with 2 mol of KOH

for 0.3 mol of H2SO4, 0.6 mol of KOH is required

But we have 0.196 mol of KOH

so, KOH is limiting reagent

According to balanced equation

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

= (1/2)*0.196

= 9.8*10^-2 mol

mol of H2SO4 remaining = mol initially present - mol reacted

mol of H2SO4 remaining = 0.3 - 9.8*10^-2

mol of H2SO4 remaining = 0.202 mol

Total volume = 0.750 L + 0.700 L = 1.45 L

Use:

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

= 0.202 mol / 1.45 L

= 0.139 M

Answer: 0.139 M