b Describe what you could uU U liitin repeat it Calculations

b) Describe what you could uU U liitin repeat it. Calculations. Be sure to show all calculations. Show the correct number of significant figures in your answers. 1. Sodium bicarbonate (NaHCO3) has a solubility of 7.00 g in 100 mL of water at 0°C. a. Calculate the number of grams of NaHCO3 present in 1.00 mL of saturated aqueous sodium bicarbonate solution at 0°C. b. Calculate the number of moles of NaHCO3 present in 1.00 mL of saturated aqueous sodium bicarbonate solution at 0°C. c. One mole = 1000 millimoles. Calculate the number of millimoles of NaHCO3 present in 1.00 mL of saturated aqueous sodium bicarbonate solution at 0°C. 2. Perform the following calculations based on benzoic acid. a. Calculate the number of moles of benzoic acid in 0.120 grams of benzoic acid. b. Calculate the number of millimoles of benzoic acid in 0.120 grams of benzoic acid. c. Assume that one mole benzoic acid is converted to one mole of sodium benzoate with the help of a base. Calculate the millimoles of sodium benzoate present from reaction of 0.120 grams of benzoic acid with base. 3. Concentrated HCl is 12 M which is 12 moles HCl per liter or 12 millimoles HCl per milliliter. a. Calculate the sum of answers lc and 2c. This is the maximum total number of millimoles of bases from an extraction process such as the one we conducted that would need to receive protons from an acid such as HCl in a neutralization. b. Calculate the number of milliliters of concentrated HCl needed to put protons on the millimoles of bases you calculated in question 3a. c. In our extraction process, 0.5 mL of concentrated HCl was used to put protons on all the bases we had present to neutralize them. Compare 0.5 mL to your answer to 3b. Was 0.5 mL enough HCI? Explain.

Solution

1)

a)

solubility of Na2CO3 = 7.00 g in 100 mL water

7.00 g NaHCO3 --------------> 100 ml water

         ??             ---------------->   1 ml water

mass of NaHCO3 = 1 x 7 / 100

mass of NaHCO3 present in 1 ml = 0.0700 g

b)

moles of NaHCO3 = 0.0700 / 84

moles of NaHCO3 = 8.33 x 10^-4 mol

c)

number of millimoles = 0.833