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A buffer solution contains 0.229 M ammonium chloride and 0.457 M ammonia. If 0.0568 moles of...

A buffer solution contains 0.229 M ammonium chloride and 0.457 M ammonia. If 0.0568 moles of hydroiodic acid are added to 250A buffer solution contains 0.443 M ammonium chloride and 0.314 M ammonia. If 0.0313 moles of potassium hydroxide are added toBase Formula Кь Ammonia NH3 1.8x10-5

A buffer solution contains 0.229 M ammonium chloride and 0.457 M ammonia. If 0.0568 moles of hydroiodic acid are added to 250 mL of this buffer, what is the pH of the resulting solution ? (Assume that the volume does not change upon adding hydroiodic acid) pH = Submit Answer Retry Entire Group 9 more group attempts remaining
A buffer solution contains 0.443 M ammonium chloride and 0.314 M ammonia. If 0.0313 moles of potassium hydroxide are added to 225 mL of this buffer, what is the pH of the resulting solution ? (Assume that the volume does not change upon adding potassium hydroxide) pH =
Base Formula Кь Ammonia NH3 1.8x10-5
0 0
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Answer #1

1)

mol of HI added = 0.0568 mol

NH3 will react with H+ to form NH4+

Before Reaction:
mol of NH3 = 0.457 M *0.25 L
mol of NH3 = 0.1143 mol

mol of NH4+ = 0.229 M *0.25 L
mol of NH4+ = 0.0573 mol

after reaction,
mol of NH3 = mol present initially - mol added
mol of NH3 = (0.1143 - 0.0568) mol
mol of NH3 = 0.0575 mol

mol of NH4+ = mol present initially + mol added
mol of NH4+ = (0.0573 + 0.0568) mol
mol of NH4+ = 0.1141 mol
since volume is both in numerator and denominator, we can use mol instead of concentration
Kb = 1.8*10^-5

pKb = - log (Kb)
= - log(1.8*10^-5)
= 4.745

use:
pOH = pKb + log {[conjugate acid]/[base]}
= 4.745+ log {0.1141/5.745*10^-2}
= 5.043

use:
PH = 14 - pOH
= 14 - 5.0425
= 8.9575
Answer: 8.96

2)

mol of KOH added = 0.0313 mol

NH4+ will react with OH- to form NH3

Before Reaction:
mol of NH3 = 0.443 M *0.225 L
mol of NH3 = 0.0997 mol

mol of NH4+ = 0.314 M *0.225 L
mol of NH4+ = 0.0707 mol

after reaction,
mol of NH3 = mol present initially + mol added
mol of NH3 = (0.0997 + 0.0313) mol
mol of NH3 = 0.131 mol

mol of NH4+ = mol present initially - mol added
mol of NH4+ = (0.0707 - 0.0313) mol
mol of NH4+ = 0.0394 mol
since volume is both in numerator and denominator, we can use mol instead of concentration
Kb = 1.8*10^-5

pKb = - log (Kb)
= - log(1.8*10^-5)
= 4.745

use:
pOH = pKb + log {[conjugate acid]/[base]}
= 4.745+ log {3.935*10^-2/0.131}
= 4.222

use:
PH = 14 - pOH
= 14 - 4.2225
= 9.7775
Answer: 9.78

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