Question

A volume of 30.0mL of a 0.810\it M \rm HNO_3 solution is titrated with 0.340\it M...

A volume of 30.0mL of a 0.810\it M \rm HNO_3 solution is titrated with 0.340\it M \rm KOH. Calculate the volume of \rm KOH required to reach the equivalence point
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Concepts and reason

The concept used to solve this problem stoichiometry of balance reaction.

Stoichiometry of balance reaction indicates the relationship between reactant and products of any chemical reaction.

Fundamentals

A balanced chemical equation is an equation that contains same number of atoms of each element on each side of reaction.

Acid-base neutralization reaction:

When an acid reacts with a base to give water and an ionic salt, the reaction is known as Acid-base neutralization reaction.

For example:

Ca(OH), (aq) + 2HNO, (aq) → 2H20 (1) + Ca(NO2)2(aq)

Molarity:

The number of moles of solute per liter of solution is known as molarity of the solution. It symbolized by . Molarity of any solution is a ratio of moles of solutes per liter of solution. It expressed by the following formula:

moles of solutes
Molarity (M)=
liter of solution
mol
= mol -

Equivalent point:

Equivalent point is a point of acid-base titration where acid and base present in the stoichiometrically equivalent amount.

Indicator:

Indicator of acid-base titration is a compound that changes color according to changes of solution. In general the color of indicator of acid-base titration changes when the acidic solution becomes basic or basic solution becomes acidic.

End point:

End point of acid –base titration is a point at which the acidic solution becomes basic or basic solution becomes acidic means at this point the indicator changes its color. In general to determinate the end point we use an indicator in acid-base titration.

To calculate the number of moles uses the following expression:

Moles of solutes Molarity (mole L )x liter of solution or volume
Number of moles =
_Amount in g
Molar mass

The balance reaction between HNO,
and is as follows:

HNO3(aq) + KOH(aq) → KNO,(aq) + H2O(1)
Reactants
Products

According to the problem; there are 30.0 mL
of a 0.810 M HNO
solution is titrated with0.340 M KOH
.


‎ First calculate the moles of HNO,
as follows:

Moles of HNO, = Molarity (mole L-) liter of solution or volume
1.00 L
= 0.810 mole L x 30 mL *
1000 mL
= 0.0243 moles HNO,

In this reaction 1.00 mole of HNO,
reacted with 1.00 mole of KOH
Therefore 0.0243 moles HNO,
will react with the following number of :

1.00 mole KOH
0.0243 moles HNO, X
= 0.0243 moles KOH
1.00 moles HNO,

Now calculate the volume of as follows:

Liter of solution or volume =
Moles of solutes
Molarity (M)
0.0243 moles KOH
Volume of HNO,
0.340 mole L KOH
= 0.0715 L KOH

Now change the unit of volume from L to mL as follows:

1.0 L = 1000 mL
0.0715 L KOH x 1000 ml = 71.5 mL KOH
1.0L

Ans:

Thus, to reach equivalence point, 71.5 mL of 0.340 M KOH
is required.

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