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1.a 1.80 l water sample is thought to contain cadmium ions. if 26.00 g cadmium phosphate (527.18 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 cd2 in the original water sample?



Answer :

In this particular problem, the molar concentration of cd2 in the original water sample 0.0822 M.

Explain how molar concentration comes 0.0822M

Considering:-

Normality= Molarity*n-factor

So, Given that:-

Molarity of Na3po4  = 1.28 M

n-factor of Na3po4  = 3

So,  

Normality of Na3po4 = 3*1.28 N = 3.84 N

Considering:-

At equivalence point

Gram equivalents of  Cd2+ = Gram equivalents of Na3po4

Given that:

Normality Na3po4 = 3.84N

VolumeNa3po4 = 77.1 ml

Volume cd2+ = 1800 ml

So,  

Normality cd2+ * 1800 = 3.84*77.1

Normality cd2+ = 296.064/ 1800 = 0.16448N

So,.

Molarity = Normality/ n-factor

           = 0.16448/ 2

           = 0. 0822M

To learn more about molar concentration.

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