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

Respuesta :

Answer:

e. 0.0822 M

Explanation:

Considering:-

[tex]Normality=Molarity\times {n-factor}[/tex]

So, Given that:-

Molarity of [tex]Na_3PO_4[/tex] = 1.28 M

n-factor of [tex]Na_3PO_4[/tex] = 3

So,  

Normality of [tex]Na_3PO_4[/tex] = 3*1.28 N = 3.84 N

Considering:-

At equivalence point

Gram equivalents of [tex]Cd^{2+}[/tex] = Gram equivalents of [tex]Na_3PO_4[/tex]

So,

[tex]Normality_{Cd^{2+}}\times Volume_{Cd^{2+}}=Normality_{Na_3PO_4}\times Volume_{Na_3PO_4}[/tex]

Given  that:

[tex]Normality_{Na_3PO_4}=3.84\ N[/tex]

[tex]Volume_{Na_3PO_4}=77.1\ mL[/tex]

[tex]Volume_{Cd^{2+}}=1.80\ L=1800\ mL[/tex]

So,  

[tex]Normality_{Cd^{2+}}\times 1800=3.84\times 77.1[/tex]

[tex]Normality_{Cd^{2+}}=\frac{296.064}{1800}=0.16448\ N[/tex]

Also,

n-factor of [tex]Cd^{2+}[/tex] = 2

So, [tex]Molarity=\frac{Normality}{n-factor}=\frac{0.16448}{2}=0.0822\ M[/tex]

Hence, e is the answer.

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