Question

a) What is the resulting concentration (M) of the solution containing 0.25 mL of 6M HCl and 12.50 mL distilled water? (0.0.75 pt.)

b) What is the pH of this solution containing the HCl? (0.05 pt.)

c) If 0.25 mL of 6M HCl is added in 4 mL of 0.2 M ClO- buffer solution, what is the final pH of the buffer (Ka HClO = 3.0 x 10-8)? (0.075 pt)

d) If the Ka of the HClO is 3.0 x 10-8, what was the pH of the orginal buffer? (0.0.75 pt.) Show your calculations.

Answer 1

a)

mol of HCl = MV = (0.25*10^-3)(6) = 0.0015

Vtotal = 0.25+12.50 = 12.75 mL

[H+} = mol/V =0.0015/(12.75 *10^-3) = 0.1176 M

b)

pH = -log(H+) = -log(0.1176 = 0.9295

c)

pH = pKa + log(ClO-/HClO)

pKa = -log(3*10^-8) = 7.52

mmol of ClO- formed = 0.2*4 = 0.8

mmol of H+ added = MV = 6*0.25 =1.5

after reaction

HClO formed = 0.8

mmol of H+ left = 1.8-0.8 = 0.7

Vtotal = 0.25+4 = 4.25

[H+] = 0.7/4.25) = 0.164

ph= -log(0.164

ph = 0.7851

d)

find pH of buffer- --> requires HClO initial