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Oliga [24]
4 years ago
5

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.
Chemistry
2 answers:
Nadya [2.5K]4 years ago
8 0

Answer:

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.

Sergio039 [100]4 years ago
6 0

Answer:

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

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\times 0.590 M\times 90.00=1\times 0.350 M\times V_2

V_2=\frac{1\times 0.590 M\times 90.0 mL}{1\times 0.350 M}=151 .7 mL\approx 152 mL

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

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