Question

(17.2b.5)

Determine the pH change when 0.077 mol HCl is added to 1.00 L of a buffer solution that is 0.438 M in HClO and 0.266 M in ClO^-.

pH after addition − pH before addition = pH change =

Answer 1

Ka of HClO = 3.0*10^-8

Lets calculate the initial pH

Ka = 3*10^-8

pKa = - log (Ka)

= - log(3*10^-8)

= 7.523

use:

pH = pKa + log {[conjugate base]/[acid]}

= 7.523+ log {0.266/0.438}

= 7.306

Lets calculate the final pH

mol of HCl added = 0.077 mol

ClO- will react with H+ to form HClO

Before Reaction:

mol of ClO- = 0.266 M *1.0 L

mol of ClO- = 0.266 mol

mol of HClO = 0.438 M *1.0 L

mol of HClO = 0.438 mol

after reaction,

mol of ClO- = mol present initially - mol added

mol of ClO- = (0.266 - 0.077) mol

mol of ClO- = 0.189 mol

mol of HClO = mol present initially + mol added

mol of HClO = (0.438 + 0.077) mol

mol of HClO = 0.515 mol

Ka = 3*10^-8

pKa = - log (Ka)

= - log(3*10^-8)

= 7.523

since volume is both in numerator and denominator, we can use mol instead of concentration

use:

pH = pKa + log {[conjugate base]/[acid]}

= 7.523+ log {0.189/0.515}

= 7.088

pH change = |final pH - initial pH|

= 7.088 - 7.306

= -0.218

Answer: -0.218