Question

QUESTION
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How many ml of 0.1 M HCL are required to react completely with 1 gram mixture of Na2Co3 and NaHCo3 containing equimolar amounts of both?






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Harsh & Lavanya ​

Answer 1

Answer:

158 mL

Hence, 158 mL of 0.1 M of HCl is required to react completely with 1 g mixture of Na2CO3 and NaHCO3, containing equimolar amounts of both

Answer 2

$\huge\pink{SoluTioN}$

Let the mixture contains$x g$ of sodium carbonate and $1−x g$ of sodium bicarbonate.

The molar masses of sodium carbonate and sodium bicarbonate are $106 g/mol \:and\: 84 g/mol$respectively .

The number of moles of sodium carbonate and sodium bicarbonates are $x/106 \:and \:1-x/84.$

Since, it is an equilmolar mixture,

$x/106 = 1-x/84$

$84x=106−106x$

$190x=106$

$x=0.5579$

Number of moles of sodium carbonate $=0.5579/106=0.005263$

Number of moles of sodium hydrogen carbonate=

$1-0.5579/84 =0.005263$

One mole of sodium carbonate will react with 2 moles of HCl and 1 mole of sodium bicarbonate will react with 1 mole of HCl.

Total number of moles of HCl that will completely neutralize the mixture $=2×0.005263+0.005263=0.01578\: moles$

Volume of 0.1 M HCl required $=0.01578/0.1=</p><p> =0.158L=158 mL$

$\huge\purple{AnsWeR}$

158 mL

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