In: Chemistry
A buffer solution consists of .195 M HCN and .22 M NaCN.
a) what is the pH of this buffer solution?
b) you add 50 mL of .15 M HCl to 300.0 mL of this buffer solution. What will the pH be?
c) You add 20.0 mL of .26 M NaOH to 150.0 mL of this buffer solution. What will the pH be?
a) pH = pKa + log [CN-]/[HCN]
pKa of HCN = 9.31
pH = 9.31 + log [0.22] / [0.195]
pH = 9.36
b) initially
millimoles of NaCN = 300 x 0.22 = 66.0
millimoles of HCN = 300 x 0.195 = 58.5
millimoles of HCl added = 50 x 0.15 = 7.50
after HCl added
millimoles of HCN = 58.5 + 7.50 = 66.0
millimoles of NaCN = 66.0 - 7.50 = 58.50
[NaCN] = 58.50 / 350 = 0.167 M
[HCN] = 66.0 / 350 = 0.188 M
pH = 9.31 + log [0.167] / [0.188]
pH = 9.26
c) millimoles of HCN = 150 x 0.195 = 29.25
millimoles of NaCN = 150 x 0.22 = 33.0
millimoles of NaOH = 20.0 x 0.26 = 5.20
after NaOH added
millimoles of HCN = 29.25 - 5.20 = 24.05
millimoles of NaCN = 33.0 + 5.20 = 38.20
[HCN] = 24.05 / 170 = 0.141 M
[NaCN] = 38.20 / 170 = 0.224 M
pH = 9.31 + log [0.224] / [0.141]
pH = 9.51