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Nucleophilic Addition at the Carbonyl

Given two mix...

Question

Given two mixtures: (A) $N a O H$ and $N a_{2} C O_{3}$ and $(B)$ $N a H C O_{3}$ and $N a_{2} C O_{3}$ .
$100$ mL of mixture (A) required $a$ mL and $b$ mL of $1 M$ $H C l$ in separate titration using phenolphthalein and methyl orange indicators respectively while $100$ mL of mixture (B) required $x$ mL and $y$ mL of same $H C l$ solution is separate titration using the same indicators.

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Solution

(A) Mixture $A$ : $(N a O H + N a_{2} C O_{3})$
$p m E q q m E q - ----------- -$
$100 m L$
With phenolphthalein: $(N a O H$ reacts completely and $1 / 2 m E q$ of $N a_{2} C O_{3}$ reacts $)$
$p + \frac{q}{2} = a \times 1$ $. . . . (i)$
with methyl orange (both react completely):
$p + q = b \times 1$ $. . . . (i i)$
Solve: $\frac{q}{2} = (b - a), q = 2 (b - q)$
mmoles of $N a_{2} C O_{3} (q) = 2 (b - a)$
mmoles of $N a_{2} C O_{3} (q) = \frac{2 (b - q)}{2} = (b - a) m m o l$
(B) Mixture $B$ : $(N a H C O_{3} + N a_{2} C O_{3})$
$r m E q s m E q - ----------- -$
$100 m L$
With phenolphthalein: $(N a H C O_{3}$ does not react $) N a_{2} C O_{3}$ react half $)$
$∴ \frac{s}{2} = x \times 1 m E q$ of $N a_{2} C O_{3}$
$s = 2 x m E q$ of $N a_{2} C O_{3} = \frac{2 x}{2}$
$= x$ mmol of $N a_{2} C O_{3}$
With methyl orange (both reacts completely):
$r + s = y \times 1 m E q$
$r + 2 x = y$
(D)r=(y-2x) mEq = $m m o l e s o f$ NaHCO_{3} $(C)$ NaOH$ in Mixture A:
Solve equations $(i)$ and $(i i)$
$2 p + \frac{3}{2} q = a + b$
Substitute the value of $q = 2 (b - a)$
$2 p + \frac{3}{/ 2} \times / 2 (b - a) = a + b$
$2 p = a + b - 3 b + 3 a = 4 a - 2 b$
$p = (2 a - b) M E q =$ mmoles of $N a O H$

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Similar questions

Q. In the mixture of

N a H C O_{3} + N a_{2} C O_{3}

H C l

x

mL with phenolphthalein indicator and then

y

mL with methyl orange indicator in same titration. Hence, volume of

H C l

for complete reaction of

N a_{2} C O_{3}

Q. In the mixture of

(N a H C O_{3} + N a_{2} C O_{3})

volume of HCl required is x mL with phenolphthalein indicator and y mL with methyl orange indicator in the same titration. Hence, volume for complete reaction of

N a_{2} C O_{3}

Q. A solution contains a mixture of

N a_{2} C O_{3}

N a O H .

Using phenolphthalein as indicator

35 m L

of mixture required

20 m L

0.80 N H C l

for the end point. With methyl orange,

35 m L

of solution required

37.5 m L

H C l

for the end point. What is the weight of

N a_{2} C O_{3}

N a O H

respectively in the mixture?

25

mL solution containing a mixture of

N a_{2} C O_{3}

N a O H

19.5

0.995 N

H C l

for the phenolpthalein end-point and in a separate titration,

25

mL of the same mixture required

25.9

H C l

for methyl orange end-point. The concentration (in g/L) of

N a_{2} C O_{3}

in the mixture is

9 X

. Find the value of

X

.

25 m L

of a mixture of

N a O H

N a_{2} C O_{3}

when titrated with

N / 10 H C l

using phenolphthalein indicator required

25 m L H C l

. The same volume of mixture when titrated with

N / 10 H C l

using methyl orange indicator required

30 m L

H C l

. Calculate the amount of

N a_{2} C O_{3}

N a O H

in one litre of this mixture.

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