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

Answer:

e. 0.0822 M

Explanation:

Considering:-

$Normality=Molarity\times {n-factor}$

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,

$Normality_{Cd^{2+}}\times Volume_{Cd^{2+}}=Normality_{Na_3PO_4}\times Volume_{Na_3PO_4}$

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,  

$Normality_{Cd^{2+}}\times 1800=3.84\times 77.1$

$Normality_{Cd^{2+}}=\frac{296.064}{1800}=0.16448\ N$

Also,

n-factor of $Cd^{2+}$ = 2

So, $Molarity=\frac{Normality}{n-factor}=\frac{0.16448}{2}=0.0822\ M$

Hence, e is the answer.