Question

Determine the pH change when 0.081 mol HI is added to 1.00 L of a buffer solution that is 0.349 M in HCN and 0.342 M in CN-.

Answer 1

we know that

for buffers

pH = pKa + log [base / acid]

in this case

pH = pKa + log [ CN- / HCN]

so initially

pH = pKa + log [0.342 / 0.349]

pH = pKa - ( 8.8 x 10-3)

now

0.081 mol HI is added

we know that

moles = concentration x volume (L)

so

initial moles of HCN = 0.349 x 1 = 0.349

initial moles of CN- = 0.342 x 1 = 0.342

now

the reaction is

CN- + H+ --> HCN

we can see that

moles of CN- reacted = moles of HI added = 0.081

moles of HCN produced = moles of HI added = 0.081

so

finally

moles of CN- = initial - reacted = 0.342 - 0.081 = 0.261

moles of HCN = initial + produced = 0.349 + 0.081= 0.43

now

pH = pKa + log [CN-/HCN]

final pH = pKa + log [ 0.261 /0.43]

final pH = pKa - 0.2168

so

pH change = final - initial

pH change = (pKa - 0.2168) - (pKa - 0.008)

pH change = -0.208

so

the pH is decreased by 0.208 units