Question

A 1.80 L water sample is thought to contain cadmium ions. If 26.00 g cadmium phosphate (527.2 g/mol) precipitates when 77.1 mL of a 1.28 M sodium phosphate solution is added to the water sample, what is the molar concentration of Cd in the original water sample?

a. 0.148 M
b. 0.164 M
c. 0.0548 M
d. 0.0493 M
e. 0.0822 M

Answer 1

e. 0.0822 M

Step-by-step explanation:

Considering:-

`Normality=Molarity* {n-factor}`

So, Given that:-

Molarity of
`Na_3PO_4` = 1.28 M

n-factor of
`Na_3PO_4` = 3

So,

Normality of
`Na_3PO_4` = 3*1.28 N = 3.84 N

Considering:-

At equivalence point

Gram equivalents of
`Cd^(2+)` = Gram equivalents of
`Na_3PO_4`

So,

`Normality_(Cd^(2+))* Volume_(Cd^(2+))=Normality_(Na_3PO_4)* Volume_(Na_3PO_4)`

Given that:

`Normality_(Na_3PO_4)=3.84\ N`

`Volume_(Na_3PO_4)=77.1\ mL`

`Volume_(Cd^(2+))=1.80\ L=1800\ mL`

So,

`Normality_(Cd^(2+))* 1800=3.84* 77.1`

`Normality_(Cd^(2+))=(296.064)/(1800)=0.16448\ N`

Also,

n-factor of
`Cd^(2+)` = 2

So,
`Molarity=(Normality)/(n-factor)=(0.16448)/(2)=0.0822\ M`

Hence, e is the answer.