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Hello, I have been struggling with this problem on my homework.

Determine the oxidation state for each atom in the following compounds: (7 pts) HPO4-2 Н Cl2 Ci Na2SO4 Na 3. I have a stock s

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Answer #1

HPO42-

Since the charge of ion is -2 we can equate this
to sum of individual oxidation numbers of each atom
from rule of oxidation numbers, H and O has following
values,using it we can get number for P
Oxidation number for H = +1
Oxidation number for O = -2.
let x be oxidation number of P
then we can write

+1 + x + [4*(-2)] = -2
1+x-8 =-2
x-7=-2
x= -2 +7 =+5
Thus oxidation number of P = +5
b.
Cl2
According to rule of oxidation number
oxidation number for a neutral molecule =0
Thus Cl has oxidation number =0
***************
c.
Na2SO4
1st group elements Has Oxidation number = +1
Thus
Na oxidation number = +1
O oxidation number= -2
let x be the Oxidation number for S
Since Na2SO4 is neutral molecule.sum of individual
oxidation number =0
[2*+1] +x + [4*(-2)= 0
2 +x -8 =0
-6+x=0
x= +6
Thus oxidation number for S= +6
*******************
3.
Let us use dilution formula to solve this
M1V1= M2V2*****
here
Molarity of NaCl stock M1= 11.5 M
Volume of stock NaCl - ???
required molarity M2 = 0.755 M
Required volume V2 = 250 ml
Plug in these numbers in
V1 = M2V2/M1
V1= 0.755 M * 250ml/11.5 M =16.453 ml
Thus volume required= 16.453 ml
*****************************
4.
molarity is obtained by dividing number of moles of solute[here NaCl]
by volume of solution in liters[convert given volume in to
liters is must]

Number of moles of NaCl= Given mass of NaCl/molar mass of NaCl
=46.2 g/58.44 g/mol = 0.79055 moles

volume of solution = 100ml = 0.1 L

Molarity = Number of moles of NaCl/volume of solution
= 0.79055 moles/0.1 L = 7.9055 M
********************************
Thank you.Hope it is clear now :))

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