4
N
a
+
O
2
→
2
N
a
2
O
.
By the stoichiometry of this reaction if 5 mol natrium react, then 2.5 mol
N
a
2
O
should result.
Explanation:
The molecular mass of natrium oxide is
61.98
g
⋅
m
o
l
−
1
. If
5
m
o
l
natrium react, then
5
2
m
o
l
×
61.98
g
⋅
m
o
l
−
1
=
154.95
g
natrium oxide should result.
So what have I done here? First, I had a balanced chemical equation (this is the important step; is it balanced?). Then I used the stoichiometry to get the molar quantity of product, and converted this molar quantity to mass. If this is not clear, I am willing to have another go
We are given the molar concentration of an aqueous solution of weak acid and the pH ofthe solution, and we are asked to determine the value of Ka for the acid.
The first step in solving any equilibrium problem is to write the equation for the equilibriumreaction. The ionization of benzoic acid can be written as seen in the attached image (1).
The equilibrium-constant expression is the equation number (2)
From the measured pH, we can calculate pH as seen in equation (3)
To determine the concentrations of the species involved in the equilibrium, we imagine that thesolution is initially 0.10 M in HCOOH molecules. We then consider the ionization of the acidinto H+ and HCOO-. For each HCOOH molecule that ionizes, one H+ ion and one HCOO- ionare produced in solution. Because the pH measurement indicates that [H+] = 1x 10^-4 M atequilibrium, we can construct the following table as seen in the equation number (4)
To find the value of Ka, please see equation (5):
We can now insert the equilibrium concentrations into the expression for Ka as seen in equation (6)
Therefore, 2.58x10^-4 M is the concentration of benzoic acid to have a pH of 4.0
Answer:
4.32
×
= 5.37
Explanation:
22.4 is how many liters per mole. 18.01 is the mass of H2O.