Question

What is the pH of a buffer prepared by adding 0.809 mol of the weak acid HA to 0.305 mol of NaA in 2.00 L of solution? The dissociation constant Ka of HA is 5.66×10−7.?

What is the pH after 0.150 mol of HCl is added to the buffer from Part A? Assume no volume change on the addition of the acid?

What is the pH after 0.195 mol of NaOH is added to the buffer from Part A? Assume no volume change on the addition of the base?

Answer 1

Question 1

[HA]= 0.809/ 2.00L = 0.405 M
[A-]= 0.305/ 2.00 L =0.1525 M
pKa = - log Ka =6.25
pH = 6.25 + log 0.1525/0.405 =5.825

Question 2

A- + HCl = HA + Cl-
moles A- = 0.305 - 0.150=0.155
[A-]= 0.155 / 2.00 L=0.0775 M
moles HA = 0.809 + 0.150 =0.959
[HA]= 0.959 / 2.00 L =0.480 M
pH = 6.25 + log 0.0775 / 0.480=5.45

Question 3

HA + NaOH >> A- + H2O
moles HA = 0.809 - 0.195=0.614
[HA]= 0.614 / 2.00 L =0.307 M
moles A- = 0.305 + 0.195=0.5
[A-]= 0.5 / 2.00 L =0.25
pH = 6.25 + log 0.25/ 0.307=6.16