In: Chemistry
Part A What is the pH of a buffer prepared by adding 0.809 mol of the weak acid HA to 0.608 mol of NaA in 2.00 L of solution? The dissociation constant Ka of HA is 5.66×10−7. Part B 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. (Express answer to 3 decimal places) Part C 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.
Part A What is the pH of a buffer prepared by adding 0.809 mol of the weak acid HA to 0.608 mol of NaA in 2.00 L of solution? The dissociation constant Ka of HA is 5.66×10?7.
Molarity of acid and salt are as follows:
Acid =0.809 /2.0 = 0.405 M
Salt = 0.608 /2.0 = 0.304 M
Ka of HA is 5.66×10?7
pKa = 6.25
pH = pKa + log(A-/HA)
pH =6.25 + log(0.304/0.405)
pH =6.25 + log 0.751
pH =6.25 -0.125
pH =6.125
Part B 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. (Express answer to 3 decimal places)
after addition of 0.150 mol HCl
A- decreases from 0.608 mol to 0.458 mol per 2L, so new molarity = 0.229 M
the HA increases from 0.607 mol to 0.757 mol , so new molarity =0.379 M
pH = pKa + log(A-/HA)
pH =6.25 + log(0.229/0.379)
pH =6.25 + log 0.604
pH =6.25 -0.219
pH =6.031
Part C 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.
A- increases from 0.608 mol to 0.803 mol; , so new molarity =0.803/2.0=0.402 M
HA decreases from 0.607 mol to 0.412 mol; so new molarity =0.412/2.0=0.206 M
pH = pKa + log(A-/HA)
pH =6.25 + log(0.402/0.206)
pH =6.25 + log 1.95
pH =6.25 +0.290
pH =6.54