Hi There! I had a question regarding analytical chemistry. Could you please incl

Question

Hi There! I had a question regarding analytical chemistry. Could you please include a detailed explanation and or show work for both parts of the problem. Thanks! :)

1) Suppose you have a 0.100 M NaF solution. How many mL of this solution should be added to a 50.00 mL flask to produce a diluted concentration of 1.00 × 10-2 M? How many mL ofthe 1.00 × 10-2 M solution should be added to a 50.00 mL volumetric flask to produce a diluted concentration of 1.00 x 10% M? (This process is called serial dilution. Consider the larger relative error which would be generated if you attempted to prepare the 1.00 x 10 M solution from the 0.100 M solution directly.) 2) A certain toothpaste is 0.16 wt/vol. % fluoride ion and has a density of 1.33 g/mL. From this information, calculate the mass of toothpaste which must b [T e added to a 50 mL volumetric flask to give M. (Hint: Use unit conversions to find the grains offluoride per gram toothpaste. Once I == 1 × 10 this is known, you can figure out how much toothpaste is required to deliver the needed mass of fluoride to the flask)

Explanation / Answer

Ans. #1. Given, stock [NaF] = 0.100 M

# Preparation of solution 1, 1.00 x10-2 M = 0.010 M

Using-

            C1V1 (stock soln.) = C2V2 (solution 1)

            Or, 0.100 M x V1 = 0.010 M x 50.0 mL

            Or, V1 = (0.010 M x 50.0 mL) / 0.100 M = 5.0 mL

Therefore, required volume of stock solution = 5.0 mL

Preparation: Transfer 5.0 mL stock solution to 50.0 mL-flask. Make the final volume upto the mark with distilled water. It is the desired solution 1.

# Preparation of solution 2, 1.00 x10-3 M = 0.001 M

Using-

            C1V1 (solution 1) = C2V2 (solution 2)

            Or, 0.010 M x V1 = 0.001 M x 50.0 mL

            Or, V1 = (0.001 M x 50.0 mL) / 0.010 M = 5.0 mL

Therefore, required volume of stock solution = 5.0 mL

Preparation: Transfer 5.0 mL of solution 1 to 50.0 mL-flask. Make the final volume upto the mark with distilled water. It is the desired solution 2.

#2. Required moles of F- in solution = Molarity x Volume of solution in liters

                                                = 1.0 x 10-4 M x 0.050 L

                                                = 5.00 x 10-6 moles

Required mass of F- in solution = Required moles x Molar mass

                                                = 5.00 x 10-6 moles x (18.9984032 g/ mol)

                                                = 9.4990 x 10-5 g

# Given,          [F-] in toothpaste = 0.16 % w/v = 0.16 g F- / 100.0 mL toothpaste

                        Density of toothpaste = 1.33 g/ mL

Now,

            Volume of 1.0 g toothpaste = Mass/ density

                                                            = 1.0 g / (1.33 g/ mL)

                                                            = 0.7519 mL

F- content in 1.0 g toothpaste = Volume of 1.0 g toothpaste x [F-] in toothpaste

                                                = 0.7519 mL x (0.16 g / 100 mL)

                                                = 1.2030 x 10-3 g

So, [F-] content in toothpaste = 1.2030 x 10-3 g F- / 1.0 g toothpaste

Now, Required mass of toothpaste = Required mass of F- / [F-] content in toothpaste

                                                = 9.4990 x 10-5 g / (1.2030 x 10-3 g / 1.0 g toothpaste)

                                                = 7.8960 x 10-2 g toothpaste

                                                = 0.078960 g = 78.960 mg

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