Lab 5 Efficacy of Buffers Post Lab Questions Name Shown in t

Lab #5 Efficacy of Buffers Post Lab Questions Name Shown in the data table below with your student number, is the initial concentration of a propanoic acid solution. You will need to look up the Ka for propanoic acid in order to answer the question. a: 1.3 Y10 5 (1) Calculate the pH of the solution. (2) Next to the acid concentration in the data table is the concentration of a sodium propanoate solution. This solution is mixed with the propanoic acid. Calculate the pH of the buffered solution. (3) 0.365 g of HCI gas is bubbled into the solution of buffer in question (2) with no change in volume to the solution. What is the pH of the solution after addition of the HCI (assume 1.00 L of solution)? (4) 0.40 g of solid NaOH is added to the solution of buffer in question (2) with no change in volume to the solution. What is the pHi of the solution after addition of the NaOH (assume 1.00 L of solution)?

Solution

From the given data

(1) initial concentration of propanoic acid = 0.779 M

let x amount of acid has dissociated into H+ and propanoate ion

then,

Ka = 1.34 x 10^-5 = x^2/0.779

x = [H+] = 3.23 x 10^-3 M

pH = -log[H+] = 2.49

(2) pH of buffer (highlighted value)

Using Hendersen-Hasselbalck equation,

pH = pKa + log(propanoate/propanoic acid)

      = -log(1.34 x 10^-5) + log(0.163/0.525) = 4.36

(3) when HCl = 0.365 g/36.5 g/mol = 0.01 mol was addded to 1 L buffer

new [propanoic acid] = 0.525 M x 1 L + 0.01 mol = 0.535 mol

ne [propanoate] = 0.163 M x 1 L - 0.01 mol = 0.153 mol

pH = 4.87 + log(0.153/0.535) = 4.33

(4) when NaOH = 0.40 g/40 g/mol = 0.01 mol was addded to 1 L buffer

new [propanoic acid] = 0.525 M x 1 L - 0.01 mol = 0.515 mol

ne [propanoate] = 0.163 M x 1 L + 0.01 mol = 0.173 mol

pH = 4.87 + log(0.173/0.515) = 4.40