Question

# A buffer solution contains 0.52 mol of hypochlorous acid (HClO) and 0.39 mol of sodium hypochlorite...

A buffer solution contains 0.52 mol of hypochlorous acid (HClO) and 0.39 mol of sodium hypochlorite (NaOCl) in 3.00 L.
The Ka of hypochlorous acid (HClO) is Ka = 3e-08.

(a) What is the pH of this buffer?

pH =__________.

(b) What is the pH of the buffer after the addition of 0.21 mol of NaOH? (assume no volume change)

pH = ___________.

(c) What is the pH of the original buffer after the addition of 0.20 mol of HI? (assume no volume change)

pH = _____________.

pH = pKa + log {[ClO- ] /[HClO]}
since volume is same for both, we can use mol instead of concentration in above equation

pKa = -log Ka
= -log (3*10^-8)
= 7.523

a)
pH = pKa + log {[ClO- ] /[HClO]}
= 7.523 + log (0.39 / 0.52)
= 7.398
b)
adding NaOH, will react with HClO to form addition 0.21 mol of ClO-
mol of ClO- = 0.39 + 0.21=0.60 mol
mol of HClO = 0.52 - 0.21 = 0.31 mol

pH = pKa + log {[ClO- ] /[HClO]}
= 7.523 + log (0.60 / 0.31)
= 7.81

c)
adding HCl, will react with ClO- to form addition 0.2O mol of HClO
mol of ClO- = 0.39 - 0.20=0.19 mol
mol of HClO = 0.52 + 0.20 = 0.72 mol

pH = pKa + log {[ClO- ] /[HClO]}
= 7.523 + log (0.19 / 0.72)
= 6.944