Consider a 1.0 L buffer containing 0.109 mol L-1 HOCl and 0.095 mol L-1 OCl-. What is the pH of the solution after adding 7.2 x 10-3 mol of NaOH? Express your answer to 2 decimal places.
Ka of HOCl = 3.5*10^-8
mol of NaOH added = 0.0072 mol
HOCl will react with OH- to form OCl-
Before Reaction:
mol of OCl- = 0.095 M *1.0 L
mol of OCl- = 0.095 mol
mol of HOCl = 0.109 M *1.0 L
mol of HOCl = 0.109 mol
after reaction,
mol of OCl- = mol present initially + mol added
mol of OCl- = (0.095 + 0.0072) mol
mol of OCl- = 0.1022 mol
mol of HOCl = mol present initially - mol added
mol of HOCl = (0.109 - 0.0072) mol
mol of HOCl = 0.1018 mol
Ka = 3.5*10^-8
pKa = - log (Ka)
= - log(3.5*10^-8)
= 7.456
since volume is both in numerator and denominator, we can use mol instead of concentration
use:
pH = pKa + log {[conjugate base]/[acid]}
= 7.456+ log {0.1022/0.1018}
= 7.458
Answer: 7.46
Consider a 1.0 L buffer containing 0.109 mol L-1 HOCl and 0.095 mol L-1 OCl-. What...
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