Question

What is the final pH of the solution when 15.0 mL of 0.117 mol L-1 NaOH(aq) and 30.0 mL of 0.175 mol L-1 CGH5COOH(aq) are mixed? Assume the temperature is 25 °C. For C6H5COOH, pK-4.21 at 25°C. Give your answer accurate to two decimal places

Answer 1

Concentration of NaOH = 0.117 mol/L

Volume of NaOH solution = 15.0 ml = 15.0 L / 1000 = 0.015 L

Number of moles of NaOH = 0.117 mol/ L * 0.015 L = 0.001755 mol

Concentration of C6H5COOH solution = 0.175 mol/L

Volume of C6H5COOH solution = 30.0 ml = 30.0 L / 1000 = 0.03 L

Number of moles of C6H5COOH = 0.175 mol/L * 0.03 L = 0.00525 mol

Reaction:-

C6H5COOH (aq) + NaOH(aq) -----> C6H5COONa(aq) + H2O (l)

From reaction, 1.0 mol of NaOH reacts with 1.0 mol of C6H5COOH produces 1.0 mol of C6H5COONa so 0.001755 mol of NaOH will react with 0.001755 mol of C6H5COOH will produce 0.001755 mol of C6H5COONa.

Number of moles of C6H5COOH remaining = ( 0.00525 - 0.001755) mol = 0.003495 mol

From Henderson equation

pH = pKa + log {[Conjugate base] / [Acid]}

pH = 4.21 + log {[C6H5COONa] / [C6H5COOH]}

pH = 4.21 + log ( 0.001755 mol / 0.003495 mol)

pH = 4.21 + log 0.5021 = 4.21 + (-0.30) = 3.91

pH = 3.91