Question

Solid ${ NaHCO }_{ 3 }$ will be neutralized by ${ 40.0mL }$ of ${ 0.1M }$ ${ H }_{ 2 }{ SO }_{ 4 }$ solution. What would be the weight of solid ${ NaHCO }_{ 3 }$ in a gram?
A.${ 0.672g }$
B.${ 6.07g }$
C.${ 17g }$
D.${ 20g }$

Answer 1

Hint:
Neutralization reactions are the reactions where an acid and a base combine to form salt and water. Neutralization reactions are important to restore soil neutrality in the case of acidic and basic soils.

Complete step-by-step answer:
It is given that;
Volume of ${ H }_{ 2 }{ SO }_{ 4 }$ = ${ 40.0mL }$
Molarity of ${ H }_{ 2 }{ SO }_{ 4 }$ = ${ 0.1M }$
The required neutralization reaction is;
${ 2NaHCO }_{ 3 }{ +H }_{ 2 }{ SO }_{ 4 }{ \rightarrow Na }_{ 2 }{ SO }_{ 4 }{ +2H }_{ 2 }{ O+2CO }_{ 2 }$
The number of moles in ${ 2NaHCO }_{ 3 }$ and ${ Na }_{ 2 }{ SO }_{ 4 }$ are ${ 2 }$ and ${ 1 }$ respectively.
Therefore, the molar mass of ${ 2NaHCO }_{ 3 }$ = ${ 168g }$
And the molar mass of ${ Na }_{ 2 }{ SO }_{ 4 }$ = ${ 98g }$
Using the formula, we can calculate the number of moles;
Number of moles of ${ H }_{ 2 }{ SO }_{ 4 } = Molarity{ \times } Volume ( in mL ) $
Put the value in the above formula, we get
The number of moles of ${ H }_{ 2 }{ SO }_{ 4 }$ = ${ 40.0\times 0.1M=0.4mol }$
Also, it can be written as m-moles of ${ NaHCO }_{ 3 } when neutralized = { 4\times 2=8mol }$
Since mol = ${ w\div m\times 1000 }$ ......... (1)
where, w = weight
m= molar mass
Now, put the values in equation (1), we get
${ 8=w\div 84\times 1000 }$
${ w=84\times 8\div 1000 }$
w = ${ 0.672g }$
Hence, the weight of solid ${ NaHCO }_{ 3 }$ = ${ 0.672g }$
The correct option is A.

Additional Information:
They are also used in an acid-base titration, which is a very important way of determining how much of an acid/basic compound is there in a sample.
The acid spills that happen are fixed by neutralization.
This process also takes place when purifying water for drinking.

Note: The possibility to make a mistake is that two ${ NaHCO }_{ 3 }$ moles give 1 mole of ${ Na }_{ 2 }{ SO }_{ 4 }$ so, you have to multiply it with ${ 2 }$.