Question

A 40.0-mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate:

a.the volume required to reach the equivalence point

b.the pH after adding 15.00 mL of KOH

c.the pH at one-half the equivalence point

Answer 1

( D )

Chemical balanced equation.

HNO₂ + KOH ------> KNO₂ + H₂O

1 mole of HNO₂ react with 1 mole of KOH and produce

1 mole of KNO₂ & 1 mole of water molecule.

At equivalent point

Number of moles of = HNO₂  = NUmber of moles of KOH

A. 40.0 ml sample of 0.100 M HNO₂ is titrated with 0.200 M KOH.

( a ) The volume of KOH required to reach the quivalone point.

Number of moles of HNO₂ ( n ) = Molarity of x volumke of HNO₂ HNO₂

_{= 0.100 M X 0.0400 lit}

_{(n) KOH =} HNO₂ = 0.00400 moles

volume of KOH ( lit ) = no. of moles KOH / molarity KOH

= 0.00400 moles / 0.200 M = 0.0200 Lit ( or ) 20.0 ml.

Koly required volume