Question

A volume of 90.0 mL mL of a 0.590 M M HN O 3 HNO3 solution is titrated with 0.350 M M KOH KOH . Calculate the volume of KOH KOH required to reach the equivalence point. Express your answer to three significant figures and include the appropriate units.

Answer 1

152 ml.

Step-by-step explanation:

Given:

Volume of HNO3 = 90 ml

Molar concentration of HNO3 = 0.59 M

Molar concentration of KOH = 0.35 M

Equation of the reaction

KOH + HNO3 --> KNO3 + H2O

Number of moles of HNO3 = molar concentration × volume

= 0.59 × 0.09

= 0.0531 moles.

By stoichiometry, 1 mole of HNO3 reacts with 1 mole of KOH. Therefore,

Number of moles of KOH = 0.0531 moles.

Volume = 0.0531 ÷ 0.350

= 0.152 l

= 152 ml.

Answer 2

152 mL is the volume of KOH required to reach the equivalence point.

Step-by-step explanation:

`HNO_3(aq)+KOH(aq)\rightarrow KNO_3(aq)+H_2O(l)`

To calculate the concentration of acid, we use the equation given by neutralization reaction:

`n_1M_1V_1=n_2M_2V_2`

where,

`n_1,M_1\text{ and }V_1` are the n-factor, molarity and volume of acid which is
`HNO_3`

`n_2,M_2\text{ and }V_2` are the n-factor, molarity and volume of base which is KOH.

We are given:

`n_1=1\\M_1=0.590 M\\V_1=90.0 mL\\n_2=1\\M_2=0.350 M\\V_2=?`

Putting values in above equation, we get:

`1* 0.590 M* 90.00=1* 0.350 M* V_2`

`V_2=(1* 0.590 M* 90.0 mL)/(1* 0.350 M)=151 .7 mL\approx 152 mL`

152 mL is the volume of KOH required to reach the equivalence point.