Question
Question: Solid \({ NaHCO }_{ 3 }\) will be neutralized by \({ 40.0mL }\) of \({ 0.1M }\) \({ H }_{ 2 }{ SO }_...
Solid NaHCO3 will be neutralized by 40.0mL of 0.1M H2SO4 solution. What would be the weight of solid NaHCO3 in a gram?
A.0.672g
B.6.07g
C.17g
D.20g
Solution
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 H2SO4 = 40.0mL
Molarity of H2SO4 = 0.1M
The required neutralization reaction is;
2NaHCO3+H2SO4→Na2SO4+2H2O+2CO2
The number of moles in 2NaHCO3 and Na2SO4 are 2 and 1 respectively.
Therefore, the molar mass of 2NaHCO3 = 168g
And the molar mass of Na2SO4 = 98g
Using the formula, we can calculate the number of moles;
Number of moles of H2SO4=Molarity×Volume(inmL)
Put the value in the above formula, we get
The number of moles of H2SO4 = 40.0×0.1M=0.4mol
Also, it can be written as m-moles of NaHCO3whenneutralized=4×2=8mol
Since mol = w÷m×1000 ......... (1)
where, w = weight
m= molar mass
Now, put the values in equation (1), we get
8=w÷84×1000
w=84×8÷1000
w = 0.672g
Hence, the weight of solid NaHCO3 = 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 NaHCO3 moles give 1 mole of Na2SO4 so, you have to multiply it with 2.