Question

calculate the pH of 100 mL of a buffer that is .080 M NH4Cl and .105 M NH3 before and after the addition of 1.00mL of 6.05 M HNO3

Answer 1

we know that

for basic buffers

pOH = pKb + log [salt / acid ]

so

pOH = pKb + log [NH4Cl / NH3]

also

Kb for ammonia is 4.75

so

pOH = 4.75 + log [ 0.08 / 0.105 ]

pOH = 4.632

now

pH = 14 - pOH

so

pH = 14 - 4.632

pH = 9.368

so

the initial pH is 9.368

2)

now

HN03 is added

we know that

moles = concentration x volume (L)

so

moles of HN03 added = 6.05 x 1 x 10-3 = 6.05 x 10-3

moles of NH4Cl = 0.08 x 100 x 10-3 = 8 x 10-3

moles of NH3 = .105 x 100 x 10-3 = 10.5 x 10-3

now

the reaction is given by

NH3 + H+ ---> NH4+

we can see that

moles of NH3 reacted = moles of HN03 added = 6.05 x 10-3

moles of NH4+ formed = moles of HN03 added = 6.05 x 10-3

so

finally

moles of NH3 = 10.5 x 10-3 - 6.05 x 10-3 = 4.45 x 10-3

moles of NH4+ = 8 x 10-3 + 6.05 x 10-3 = 14.05 x10-3

now

pOH= pKb + log [NH4+ / NH3]

so

pOH = 4.75 + log [ 14.05 x 10-3 / 4.45 x10-3]

pOH = 5.25

now

pH = 14 - pOH

so

pH = 14 - 5.25

pH = 8.75

so

the pH after addition of HN03 is 8.75