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Concentration Terms

12.6g of impu...

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12.6g of impure Na2CO3 is treated with 500ml of 0.5m HCL for complete reaction .the remaining HCL is back titrated by adding 100ml of 0.5 M NaOH .calculate the percentage purity of Na2CO3

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50 mL

solution having

N a O H

N a_{2} C O_{3}

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

N a_{2} C O_{3}

?

0.2 g m

of solution of mixture of

N a O H

N a_{2} C O_{3}

and inert impurities was first titrated with phenolphthalein and

N / 10

H C l

17.5 m l

H C l

was required at the endpoint. After this methyl orange was added and

2.5 m l

H C l

was again required for next endpoint. Find out percentage of

N a O H

N a_{2} C O_{3}

in the mixture.

c m^{3}

N

HCl is titrated against 0.1

N

NaOH solution. The titration is discontinued after adding 50

c m^{3}

of NaOH. The remaining titration is completed by adding 0.5

N

KOH. The volume of KOH required for completing the titration is:

0.4 g

N a O H

N a_{2} C O_{3}

and some inert impurities were first titrated with

\frac{N}{10} H C l

using phenolphthalein as an indicator.

17.5 m L

H C l

was required at the end point. After this methyl orange was added and titrated.

1.5 m L

H C l

was required for the next end point. The weight percentage of

N a_{2} C O_{3}

in the mixture is ______. (Rounded off to the nearest integer)

Q. 1000 g of a mixture of

N a_{2} C O_{3}, N a_{2} S O_{4}

N a O H

for complete neutralisation requires 511 g of

H C l

. The same mixture when reacted with excess of

B a C l_{2}

solution, produces 466 g of white precipitate of

B a S O_{4}

. Calculate the mass percentage of

N a O H

in the mixture.
(Given Molar mass of

N a_{2} C O_{3} = 106 g, N a_{2} S O_{4} = 142 g, B a S O_{4} = 233 g, H C l = 36.5 g a n d N a O H = 40 g

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