1.Calculate how much lead nitrate [Pb(NO3)2] would be needed to make a 500 ppm Pb2+ stock...

1.Calculate how much lead nitrate [Pb(NO₃)₂] would be needed to make a 500 ppm Pb²⁺ stock solution?

Expert Solution

To determine the concentration of a substance in ppm the concentration is first expressed in terms of mol/L. In dilute solutions, ppm can be equated to mg/L as 1ppm is 1 part of substance per million parts of solution with 1mg = 10^-3g and 1mL = 10^-3L giving 1x10^-6mg/L.

Thus, to find the concentration in terms of mg/L, the molar mass of Pb²⁺ is found and then equated for the given 500ppm as 500mg/L of the metal ion. Molar mass of plumbous ion is 207.2g/mol. Therefore, 500mg = 0.5g will give 0.5/207.2 = 2.4131mmoles of lead ion in 1L of solution.

There is 1 mole of lead ion in 1 mole of lead nitrate. So, for 2.4131mmoles of lead ion, a 2.4131 millimolar solution of lead nitrate is prepared. Now, to find out the quantitiy of lead nitrate required, we multiply the concentration with the molar mass of lead nitrate as 0.0024131 x 331.2 (molar mass of lead nitrate) = 0.7992g of lead nitrate.

Therefore, we find that a 1L solution prepared with 0.7992g of lead nitrate or a 2.4131 millimolar solution of lead nitrate will contain 500ppm of Pb²⁺ ions.

For a given volume of solution with the concentration of 2.4131mmol the formula is 2.4131 x (1000/volume).

Generally, molarity is calculated as (Weight/Molar mass) x (1000/volume).

queen_honey_blossom answered 2 years ago

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