#13 A) A buffer solution contains 0.378 M ammonium bromide and 0.411 M ammonia .

Question

#13

A) A buffer solution contains 0.378 M ammonium bromide and 0.411 M ammonia.

If 0.0359 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 =

B) A buffer solution contains 0.500 M ammonium bromide and 0.397 M ammonia.

If 0.0235 moles of potassium hydroxide are added to 125 ml of this buffer, what is the pH of the resulting solution ?
(Assume that the volume does not change upon adding potassium hydroxide)

pH =

Explanation / Answer

A)

we know that

moles = molarity x volume (L)

moles of Nh4+ = 0.378 x 0.15 = 0.0567

moles of NH3 = 0.411 x 0.15 = 0.06165

now

0.0359 moles of HI is added

the reaction is

H+ + NH3 ----> NH4+

so

from the above reaction

moles of NH3 reacted = moles of H+ added = 0.0359

new moles of NH3 = 0.06165 - 0.0359 = 0.02575

moles of NH4+ formed = moles of NH3 reacted = 0.0359

new moles of NH4+ = 0.0567 + 0.0359 = 0.0926

now we know that for a basic buffer

pOH = pKb + log [ conjugate acid / base ]

pOH = pKb + log [ NH4+ / NH3]

conc = moles / volume

as the final volume is same for both , they cancel out

so

ratio of conc = ratio of moles

aslo pKb for Nh3 is 4.75

so

pOH = 4.75 + log [ 0.0926 / 0.02575]

pOH= 5.30

we know that

pH = 14 - pOH

pH = 14 - 5.31

pH = 8.69

so the pH of resulting solution is 8.69

B)

moles = molarity x volume (L)

moles of Nh4+ = 0.5 x 0.125 = 0.0625

moles of NH3 = 0.397 x 0.125 = 0.049625

now

0.0235 moles of KOH is added

the reaction is

NH4+ + OH- ----> NH3 + H20

so

from the above reaction

moles of NH4+ reacted = moles of OH- added = 0.0235

new moles of NH4+ = 0.0625 - 0.0235 = 0.039

moles of NH3 formed = moles of NH4+ reacted = 0.0235

new moles of NH3 = 0.049625 + 0.0235 = 0.073125

now we know that for a basic buffer

pOH = pKb + log [ conjugate acid / base ]

pOH = pKb + log [ NH4+ / NH3]

conc = moles / volume

as the final volume is same for both , they cancel out

so

ratio of conc = ratio of moles

aslo pKb for Nh3 is 4.75

so

pOH = 4.75 + log [ 0.039 / 0.073125]

pOH= 4.477

we know that

pH = 14 - pOH

pH = 14 - 4.477

pH = 9.523

so the pH of resulting solution is 9.523

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