1. Suppose that you have 0.500 L of each of the following solutions, and an unli

Question

1. Suppose that you have 0.500 L of each of the following solutions, and an unlimited supply of water. (Note: C9H7NHBr is a salt containing the ions C9H7NH+ and Br? and C9H7N is quinoline, an organic base with pKb = 6.24 at 298 K. If you like, you may represent C9H7NH+ as HB+ and C9H7N as B.) 0.113 mol L?1 C9H7NHBr (aq) 0.104 mol L?1 HBr(aq) 0.122 mol L?1 NaOH(aq) (1a) (6 marks) Provide simple instructions for preparing 1.00 L of a solution having pH = 7.00 at 298 K. Your instructions should include the volumes of the solutions required. (1b) (4 marks) What is the buffer capacity of the resulting solution? (The buffer capacity is the number of moles of NaOH that must be added to 1.0 L of solution to raise the pH by one unit.)

Explanation / Answer

Remember the definition of molarity: Molarity = moles / liters 1) 0.997 kg NaNO3 Convert grams to moles by dividing by the molar mass of KNO3 997 g / 85.0 g/mol = 11.73 moles molarity = moles / liters = 11.73 moles / 125 L = 0.0938 M 2) 0.50 moles of KBr / 0.250 L = 2.00 M 3) 0.500 M NaOH means 0.500 moles NaOH in 1 liters of solution 0.500 mole NaOH x 40.0 grams/mole = 20.0 grams NaOH to prepare 1 liter. If you want 10.0 L, you need 10 x 20.0 g or 200 grams. If you want 10.0 mL (0.010 L) you need 0.010 x 20.0 g or 0.200 g NaOH. (You need 10.0 mL or L - you left out the units.) 5. moles For example, suppose you diluted 5.00 mL of 2 M NaOH to a total volume of 25.00 mL. The moles of NaOH did not change. The molarity changed. 0.00500 L x 2 mol/L = 0.0100 moles of NaOH After dilution, you still have 0.0100 moles of NaOH but not the total solution is 25 mL (0.025 L) and the molarity will be 0.0100 mole/0.025 L = 0.40 M

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