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Standard XII

Chemistry

Salt of Strong Acid and Weak Base

5.00 mL of 0....

Question

$5.00 mL$ of $0.10 M$ oxalic acid solution taken in a conical flask is titrated against $N a O H$ from a burette using phenolphthalein indicator. The volume of $N a O H$ required for the appearance of permanent faint pink color is tabulated below for five experiments. What is the concentration, in molarity, of the $N a O H$ solution?

Exp. No.

Vol. of NaOH (mL)

1

12.5

2

10.5

3

9.0

4

9.0

5

9.0

0.111

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0.11

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0.10

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Solution

$H_{2} C_{2} O_{4} + 2 N a O H 9 mL \to N a_{2} C_{2} O_{4} + 2 H_{2} O$
By equivalence concept,
$Milliequivalent of H_{2} C_{2} O_{4} = Milliequivalent of N a O H$
$5 \times 0.1 \times 2 = 9 \times x x = \frac{1}{9} M x = 0.11 M$

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Similar questions

5.00 mL

0.10 M

oxalic acid solution taken in a conical flask is titrated against

N a O H

from a burette using phenolphthalein indicator. The volume of

N a O H

required for the appearance of permanent faint pink color is tabulated below for five experiments. What is the concentration, in molarity, of the

N a O H

solution?

Exp. No.	Vol. of NaOH (mL)
1	12.5
2	10.5
3	9.0
4	9.0
5	9.0

Q. The strength of an aqueous NaOH solution is most accurately determined by titrating:
(Note: consider that an appropriate indicator is used)

Q. Consider titration of

N a O H

solution versus 1.25

M

oxalic acid solution. At the end point following burette readings were obtained.

(i) 4.5 mL (i i) 4.5 mL (i i i) 4.4 mL (i v) 4.4 mL (v) 4.4 mL

If the volume of oxalic acid taken was

10.0 mL

. Then the molarity of the

N a O H

solution is

M

. (Rounded-off to the nearest integer)

Q. In an experiment,

20 m L

of decinormal

H C l

solution was added to

10 m L

of a decinormal

A g N O_{3}

solution.

A g C l

was precipitated out and the excess acid was titrated against a decinormal

N a O H

solution. Volume of

N a O H

required for this back titration is:

50 mL

solution having

N a O H

and

N a_{2} C O_{3}

require,

0.5 M 10 mL H C l

for complete titration with phenolphthalein indicator and another

50 mL

solution require

0.5 M, 12 mL H C l

solution in methyl orange indicator then what is the ratio of number of mole of

N a O H

and

N a_{2} C O_{3}

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H2C2O4+2NaOH9 mL→Na2C2O4+2H2O By equivalence concept, Milliequivalent of H2C2O4=Milliequivalent of NaOH 5×0.1×2=9×xx=19 Mx=0.11 M

$H_{2} C_{2} O_{4} + 2 N a O H 9 mL \to N a_{2} C_{2} O_{4} + 2 H_{2} O$
By equivalence concept,
$Milliequivalent of H_{2} C_{2} O_{4} = Milliequivalent of N a O H$
$5 \times 0.1 \times 2 = 9 \times x x = \frac{1}{9} M x = 0.11 M$