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?

Solutions

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 1 year ago

Next > < Previous

Related Solutions

What is the mass % of lead in lead (II) nitrate (Pb(NO3)2) rounded to three significant figures?

What is the mass % of lead in lead (II) nitrate (Pb(NO3)2) rounded to three significant figures?31.362.612579.90.331

What is the concentration (in molality) of lead nitrate (Pb(NO3)2) in a 0.798 M solution? The...

What is the concentration (in molality) of lead nitrate (Pb(NO3)2) in a 0.798 M solution? The density of the solution is 1.27 g/mL. Answer has 3 decimal places.

Suppose we have a solution of lead nitrate, Pb(NO3)2(aq). A solution of NaCl(aq) is added slowly...

Suppose we have a solution of lead nitrate, Pb(NO3)2(aq). A solution of NaCl(aq) is added slowly until no further precipitation occurs. The precipitate is collected by filtration, dried, and weighed. A total of 11.72 g of PbCl2(s) is obtained from 200.0 mL of the original solution. Calculate the molarity of the Pb(NO3)2(aq) solution.

100.0 mL of 0.500 M lead(II) nitrate (Pb(NO3)2, 331 g/mol) and 100.0 mL of 0.500 M...

100.0 mL of 0.500 M lead(II) nitrate (Pb(NO3)2, 331 g/mol) and 100.0 mL of 0.500 M sodium sulfate (Na2SO4, 142 g/mol) are mixed together. (I) Write the overall balanced equation. Include the states of each species. (II) Write the overall (or total) ionic equation for the reaction. Identify any spectator ions. Include states of each species. (III) Write the net ionic equation. Include states of each species. (IV) Determine the limiting reactant and the theoretical yield of the non-aqueous product....

An ideal Pb2+Pb2+ ion-selective electrode is moved from a 5.69×10−55.69×10−5 M Pb(NO3)2Pb(NO3)2 solution to a 3.22×10−33.22×10−3...

An ideal Pb2+Pb2+ ion-selective electrode is moved from a 5.69×10−55.69×10−5 M Pb(NO3)2Pb(NO3)2 solution to a 3.22×10−33.22×10−3 M Pb(NO3)2Pb(NO3)2 solution at 25 °C. By how many millivolts will the potential of the ion-selective electrode change when it is moved from the first solution to the second solution?

A galvanic cell consists of a lead electrode in 0.0256 M Pb(NO3)2 and a aluminum electrode...

A galvanic cell consists of a lead electrode in 0.0256 M Pb(NO3)2 and a aluminum electrode in 0.292 M Al(NO3)3. What is the cell potential (emf, in V) of this cell at 25oC?

If 550 ml of some Pb(NO3)2 solution is mixed with 500 ml of 1.30 x 10-2...

If 550 ml of some Pb(NO3)2 solution is mixed with 500 ml of 1.30 x 10-2 M NaCl solution, what is the maximum concentration of the Pb(NO3)2 solution added if no solid PbCl2 forms? (Assume Ksp = 2.00 x 10-5 M at this temperature.) Enter the concentration in M.

Lead (II) nitrate and aluminum chloride react according to the following equation: 3Pb(NO3)2 + 2AlCl3 →...

Lead (II) nitrate and aluminum chloride react according to the following equation: 3Pb(NO3)2 + 2AlCl3 → 3PbCl2 + 2Al(NO3)3. In an experiment, 8.00 g of lead nitrate reacted with 2.67 g of aluminum chloride. A. Which reactant was the limiting reagent? B. What is the theoretical yield of lead chloride? C. If 5.55 g of lead chloride is produced, what is the percent yield (actual amount of product divided by theoretical amount of product)?

Calculate the mass (in mg) of solid chromium(III) nitrate needed to make 250. mL of an...

Calculate the mass (in mg) of solid chromium(III) nitrate needed to make 250. mL of an aqueous solution with a pH of 4.00. The Ka of Cr(H2O)63+ is 1.0 x 10−4. 48 mg 3.0 mg 12 mg 6.0 mg 24 mg

Answer the following for the reaction: Pb(NO3)2(aq)+2KCl(aq)→PbCl2(s)+2KNO3(aq) How many milliliters of a 2.05 M Pb(NO3)2 solution...

Answer the following for the reaction: Pb(NO3)2(aq)+2KCl(aq)→PbCl2(s)+2KNO3(aq) How many milliliters of a 2.05 M Pb(NO3)2 solution will react with 40.5 mL of a 1.45 M KCl solution? Please show work! Thanks

Subjects