First, we need to get the concentration of [NaH2PO4]:
[NaH2PO4] =( mass / molar mass ) * volume L
when we have mass NaH2PO4 = 6.6 g & molar mass = 120g/mol & V = 0.355 L
So by substitution:
[NaH2PO4] = (6.6g / 120g/mol) * 0.355 L = 0.0195 M
then, we need to get the concentration of [Na2HPO4]:
[Na2HPO4]= (mass / molar mass ) * volume L
So by substitution:
[Na2HPO4] = (8g/ 142g/mol) * 0.355 L = 0.02 M
and when Pka of the 2nd ionization of phosphoric acid = 7.21
So by substitution in the following formula, we can get the PH:
PH = Pka + ㏒[A]/[AH]
∴PH = 7.21 + ㏒[0.02]/[0.0195]
∴ PH = 7.2
I check all of them. most of them correct but one
in question 6, the answer is the third choice. remember to find the neutrons, you take the atomic mass minus the atomic number. 38 - 18= 20
An example of a base in chemistry would be like soap or lye
Wouldn’t it be pollution? Or......
You have the stoichiometric equation. This tells you unequivocally that an
18
⋅
g
mass of water, 1 mole, reacts with a
56.07
⋅
g
mass of quicklime to form a
74.09
⋅
g
mass of slaked lime.
If you don't from where I am getting these numbers, you should know, and someone will be willing to elaborate.
Here, you have formed
6.21
⋅
m
o
l
of quicklime which requires stoichiometric lime AND water. And thus you need a mass of
6.21
⋅
m
o
l
×
18.01
⋅
g
⋅
m
o
l
−
1
water
≅
88
⋅
g
.
In practice, of course I would not weigh out this mass. I would just pour
100
−
200
⋅
m
L
of water into the lime.
