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Standard XII

Chemistry

Concentration Terms

20 mL of x ...

Question

$20$ mL of $x$ M $H C l$ neutralises completely $10$ mL of $0.1 M$ $N a H C O_{3}$ and further, $5$ mL of $0.2 M N a_{2} C O_{3}$ solution to methyl orange end-point. The value of $x$ is:

0.167 M

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0.133 M

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0.150 M

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0.200 M

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Solution

The correct option is C $0.150 M$
Methyl orange indicator gives the end point corresponding to $100 %$ of sodium carbonate.
The balanced chemical equation is as shown below:
$N a_{2} C O_{3} + 2 H C l \to 2 N a C l + H_{2} O + C O_{2}$
$20 \times x = 10 \times 0.1 + 5 \times 0.2 \times 2$
$x = 0.15 M$
Hence, the value of $x$ is $0.15 M$ .

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Similar questions

20

mL of

x M

H C l

neutralizes completely

10

mL of

0.1 M

N a H C O_{3}

solution and a further

5

mL of

0.2 M

N a_{2} C O_{3}

solution to methyl orange end point. The value of

x

is:

Q. Solution "X" contains

{N a}_{2} {C O}_{3}

and

N a H C O_{3}

. 20 ml of X when titrated using methyl orange indicator consumed 60 ml of 0.1 M

H C l

solution. In another experiment , 20 ml of X solution when titarted using phenolphthalein consumed 20 ml of 0.1 M

H C l

solution. The concentrations (in

m o l {l i t}^{- 1}

) of

{N a}_{2} {C O}_{3}

and

N a H C O_{3}

in X are respectively :

Q. For a given mixture of

N a H C O_{3}

and

N a_{2} C O_{3}

, volume of a given

H C l

required is

x

mL with phenolphthalein indicator and further

y

mL required with methyl orange indicator. Hence, volume of

H C l

for complete reaction of

N a H C O_{3}

is :

In the mixture of $N a H C O_{3}$ and $N a_{2} C O_{3}$ volume of a given HCl required is 'x' ml with phenolphthalein indicator and further 'y' ml required with methyl orange indicator. Hence volume of $H C l$ for complete reaction of $N a H C O_{3}$ is