Question

An acid solution is 0.120 M in HCl and 0.210 Min H2SO4

Can you explain this to me?

Answer 1

Volume of acidic solution = 550ml, the acidic solution contains 0.12M HCl.

moles of HCl in the acidic solution= Molarity* Volume in L= 0.12*550/1000=0.066

The reaction between HCl and KOH is HCl+ KOH------>KCl+ H2O

1 mole of HCl requires 1 mole of KOH, moles of KOH required= 0.066

Molarity of KOH given = 0.14M, Volume of KOH required ( in L) for neutralizing HCl= Moles/ Molarity =0.066/0.14=0.47L= 470ml

Moles of H2SO4 present in 550ml of 0.21M = 0.21*550/1000= 0.1155

The reaction between H2SO4 and KOH can be written as H2SO4+ 2KOH-------->K2SO4+ H2O

1 mole of H2SO4 requires 2 mole of KOH

0.1155 moles of H2SO4 requires 0.1155*2=0.231 moles of KOH

Volume of KOH requires =0.231/0.14 =1.65 L= 1650ml

Hence Volume of KOH required totally= volume requires to neutralize HCl+ Volume required to neutralize H2SO4= 470+1650 =2120ml =2.120 L