Chemistry, asked by mtotheb18, 21 days ago

Potassium hydrogen phthalate is used to standardize NaOH solution. A 0.2187 g of KHP was weighed and dissolved in 20 mL of water. If 8.85 mL of the NaOH solution was used to neutralize it, calculate the molar concentration of the NaOH solution. (molar mass of KHP = 204.22g/mol). With solutions and steps pls.

Answers

Answered by DEEPTHI09
1

Answer:

Explanation:

Your goal here is to standardize a solution of sodium hydroxide,

NaOH

, by using potassium hydrogen phthalate,

KHP

.

Your starting point here is the balanced chemical equation for this neutralization reaction

KHP

(aq]

+

NaOH

(aq]

KNaP

(aq]

+

H

2

O

(l]

The important thing to notice here is that you have a

1

:

1

mole ratio between the two reactants. This tells you that at you can reach the equivalence point by reacting equal number of moles of

KHP

and of

NaOH

.

You start with

0.5100 g

of

KHP

. To get the molar amount of acid used for the experiment, use its molar mass

0.5100

g

molar mass of KHP



1 mole KHP

204.22

g

=

0.0024973 moles KHP

So, you know that at equivalence point, the reaction will consume

0.0024973

moles of

KHP

and

0.0024973

moles of

NaOH

, since that's what the

1

:

1

mole ratio tells you.

As you know, molarity is defined as moles of solute per liters of solution.

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

c

=

n

solute

V

solution

a

a

−−−−−−−−−−−−−−−

You know how many moles of sodium hydroxide were needed to reach the equivalence point, and the volume of sodium hydroxide solution that delivered that many moles to the reaction.

This means that you can calculate its actual molarity by dividing the two - do not forget to convert the volume from milliliters to liters by using the conversion factor

1 L

=

10

3

mL

In your case, you will have

[

NaOH

]

=

0.0024973 moles

31.70

10

3

L

=

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

0.07878 M

a

a

−−−−−−−−−−−−−

As you can see, the actual molarity of the solution you prepared is relatively small compared with your target of

0.100 M

.

You can calculate the percent error by using the formula

% error

=

|

approximate value

exact value

|

exact value

×

100

In your case, you will get

% error

=

|

0.07878

0.100

|

0.100

×

100

=

21.22

%

This is characteristic of a major error in your experiment.

My guess is that you allowed too much sodium hydroxide to react with the acid, which would cause the molarity of the solution to appear to be smaller than in reality.

If you're using phenolphthalein as your indicator, an excess of sodium hydroxide would cause the solution to be a brighter shade of pink than it should be at equivalence point.

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