Question

To a 1.00 L buffer solution made of 1.95 M hypochlorous acid (HClO) and 1.03 M potassium hypochlorite (KClO) was added 0.509 moles of NaOH (assume no volume change to the solution).

What is the final pH of this buffer assuming K_a hypochlorous acid = 2.90e-8?

Answer 1

from Ka value find the pKa

pKa = -logKa = -log(2.9x10^-8) = 7.54

since the given volume is 1L moles = Molarity

moles of HClO = 1.95 mol

moles of KClO = 1.03 mol

moles of NaOH = 0.509 mol

added NaOH will react with HClO so

0.509 moles of NaOH react with 0.509 mol of HClO form the additional 0.509 mol NaClO

after addition

moles of HClO remaining = 1.95 mol - 0.509 mol = 1.44 mol

moles of ClO- = 1.03 mol + 0.509 = 1.539 mol

pH = pKa + log[ClO- / HClO]

pH = 7.54 + log[1.539 / 1.44]

pH = 7.54 + 0.03

pH = 7.57