A buffer solution contains 0.274 M hypochlorous acid and 0.430 M sodium hypochlo
ID: 1039809 • Letter: A
Question
A buffer solution contains 0.274 M hypochlorous acid and 0.430 M sodium hypochlorite.
1. If 0.0228 moles of hydroiodic acid are added to 150 mL of this buffer, what is the pH of the resulting solution ?
(Assume that the volume does not change upon adding hydroiodic acid)
pH =
2. A buffer solution contains 0.356 M ammonium bromide and 0.317 M ammonia.
If 0.0194 moles of sodium hydroxide are added to 150 mL of this buffer, what is the pH of the resulting solution ?
(Assume that the volume does not change upon adding sodium hydroxide)
pH =
Explanation / Answer
1)
Ka of hypochlorous acid (HClO) = 3.0*10^-8
mol of HI added = 0.0228 mol
ClO- will react with H+ to form HClO
Before Reaction:
mol of ClO- = 0.43 M *0.15 L
mol of ClO- = 0.0645 mol
mol of HClO = 0.274 M *0.15 L
mol of HClO = 0.0411 mol
after reaction,
mol of ClO- = mol present initially - mol added
mol of ClO- = (0.0645 - 0.0228) mol
mol of ClO- = 0.0417 mol
mol of HClO = mol present initially + mol added
mol of HClO = (0.0411 + 0.0228) mol
mol of HClO = 0.0639 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 {4.17*10^-2/6.39*10^-2}
= 7.338
Answer: 7.34
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