Question

63. You need to make a pH 10.70 buffer solution starting with 5.0 L of 2.0 M NaHCO3. You have a 2.0 L bottle of concentrated HCI (11.65 M) and 500 g bottle of solid sodium hydroxide. Show your calculations and appropriate chemical reactions to explain how you would convert the 5.0 L of 2.0 M NaHCO3 to a buffer with pH of 10.70.

Answer 1

HCO3- (aq) + OH- (aq) = CO3-2 (aq) + H2O (l)

The buffer should conatin a mixture of HCO3- and CO3-2 which can be acheived by adding NaOH to the solution of NaHCO3.

5 L of 2.0 M of NaHCO3 contains 5 * 2 mole = 10 moles of NaHCO3. Suppose you need to add x moles of NaOh that produces x moles of CO3-2. So, the mixture contains x moles of CO3-2 and (10- x) moles of HCO3-

pH = pKa + log[CO3-2]/[HCO3-]

10.70 = 10.30 + log(x/ 10-x)

So, 0.4 = log(x/ 10-x)

So, (x/ 10-x) = 10 0.40 = 2.51

So, x = 25.1 - 10x

Or, 11x = 25.1

Or, x = 25.1/ 11 = 2.28

So, you need to add 2.28 moles of NaOH to the 5 L of 2.0 M of NaHCO3

Mass of NaOH = Moles * Molar mass = 2.28 moles * 40 g/ mole = 91.2 g

Answer: You have to add 91.2 g NaOH to 5.0 L of 2.0 M solution of NaHCO3 to make the buffer of pH = 10.70.