Question

A) What is the pH after 0.28 moles of NaOH is added to a buffer containing 1.00 moles of NH₃ and 1.00 moles NH₄Cl ? K_{b NH3} = 1.8 x 10^-5

B)What is the pH after 0.36 moles of HCl is added to a buffer containing 1.00 moles of HOBr and 1.00 moles NaOBr ? K_{a HOBr} = 2.5 x 10^-9

Answer 1

1)

the reaction is

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

we can see that

moles of Nh4+ reacted = moles of NaOH added = 0.28

moles of NH3 formed = moles of NaOH added = 0.28

so

finally

moles of NH4+ = 1 - 0.28 = 0.72

moles of NH3 = 1 + 0.28 = 1.28

now

for a buffer

pOH = pKb + log [salt / base]

also

pKb = -log Kb

so

pOH = -log Kb + log [NH4+ / NH3]

pOH = -log 1.8 x 10-5 + log [ 0.72 / 1.28]

pOH = 4.5

now

pH = 14 - pOH

so

pH = 14 - 4.5

pH = 9.5

so

pH of the solution becomes 9.5

2)

now the reaction is given by

H+ + BrO- ---> HOBr

so

moles of Bro- reacted = moles of HCl added = 0.36

moles of HOBr formed = moles of HCl added = 0.36

now finally

moles of BrO- = 1 - 0.36 = 0.64

moles of HOBr = 1 + 0.36 = 1.36

now

for acid buffers

pH = -log Ka + log [salt / acid ]

so

pH = -log Ka + log [BrO- / HOBr]

pH = -log 2.5 x 10-9 + log [ 0.64 / 1.36]

pH = 8.2747

so

pH of the solution is 8.2747