How 11. You need to prepare a 0.500 liter of a solution that is 85 ppb Pb, from

Question

How 11. You need to prepare a 0.500 liter of a solution that is 85 ppb Pb, from solid Pb(NOs2 many grams of lead nitrate do you need to weigh out to prepare this solution? If a 1000. ppm lead nitrate solution is already available, how many ml of it do you need to use to prepare the 85 ppb Pb solution? (8 pts) 12. Your assignment is to spectroscopically analyze for Calcium in urine. Calcium atoms absorb visible light at 422 nm. To perform the above analysis, the following amounts of standard were pipetted from a certified 500. ppm stock Ca standard solution into 100 mL volumetric flasks and diluted to the mark with DI water: 0.0 mL, 0.500 mL, 1.00 ml, 1.500 mL, 2.00 mL. The instrument was set at 422 nm, correctly blanked on water, and these solutions were analyzed, with the following %T values: 0.0 mL 0.50 mL 1.00 mL 1.50 mL 2.00 mL 100.0%T 82.8%T 68.5%T 57196T 47.6%T Assume that the pathlength was 5.0 cm. Graph the data, giving the slope, the intercept, and the correlation coefficient, r. What is the epsilon value, e, for this absorption? You are to now finally analyze an unknown urine sample. You take 25.0 ml of the urine, dilute it in a 100.0 ml volumetric flask, and then analyze it. It produces a 96T = 92.1%. How many ppm of Ca are in the original urine unknown? (12 pts)

Explanation / Answer

11) Molar mass of Pb(NO3)2 = (1*207.2 + 2*14.0067 + 6*15.9994) g/mol = 331.2098 g/mol.

We wish to prepare 85 ppb Pb solution by dissolving solid Pb(NO3)2 in water. We know that 1 ppb = 1 µg/L; therefore, 85 ppb = 85 µg/L = (85 µg/L)*(1 g/1.0*106 µg) = 8.5*10-5 g/L.

Volume of the solution is 0.500 L; therefore, mass of solid Pb(NO3)2 required = (volume of solution)*(concentration of solution) = (0.500 L)*(8.5*10-5 g/L) = 4.25*10-5 g (ans).

We know that 1 ppm = 1 mg/L; therefore, 1000 ppm = 1000 mg/L = (1000 mg/L)*(1 g/1000 mg) = 1 g/L.

The dilution factor = (1000 ppm)/(85 ppb) = (1 g/L)/(8.5*10-5 g/L) = 11764.71.

Volume of the stock solution required = (0.500 L)*(1000 mL/1 L)/(11764.71) = 0.0425 mL (ans).

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