Question

To prepare a Cd 2+ standard, 0.280 g of cadmium metal is first dissolved in 50 mL of concentrated HCl, followed by dilution to 1.00 L. This solution is followed by another dilution that reduces the concentration by 100-fold. What is the concentration of the resulting solution? (in ppm)

Answer 1

a)

initially

mol of Cd = mass/MW = 0.28/112.4110 = 0.0024908 mol of Cd

Vinitial = 50 mL

[Cd+2] = mo/V = 0.0024908/(50*10^-3) = 0.049816 M after concentrted HCl

then

V next = 1 L

M1*V1 = M2*V2

M2 = M1*V1/V2 = 0.049816 * 50/1000 = 0.0024908 M of Cd+2 in 1 L mix

another dilution reduced to 100x

M3 = 1/100*M2 = 0.0024908/100 = 0.000024908 M

recall that

ppm = mg / kg

0.000024908 M = 0.000024908 mol / L = (0.000024908)(112.4110 ) = 0.002799 g = 0.002799*!0^3 mg = 2.799 mg

now 1 L = 1 kg approx

so

ppm = 2.799 mg / 1 kg = 2.799 ppm

answered by: anonymous