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.

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

Explanation:

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

where,

are the n-factor, molarity and volume of acid which is

are the n-factor, molarity and volume of base which is KOH.

We are given:

Putting values in above equation, we get:

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

152 ml.

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.

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