Question

Part A What is the pH of a buffer prepared by adding 0.405 mol of the weak acid HA to 0.608 mol of NaA in 2.00 L of solution? pH = 6.423 . 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 the pH numerically to three 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. Express the pH numerically to three decimal places.

Answer 1

Ans A -

[HA] = 0.405 / 2.00 = .203 M

[A-]= 0.608 / 2.00 = .304 M

pKa = 6.25

pH = 6.25 + log 0.304/ 0.203 = 6.425

Part B -

moles [A-] = .608 - .150 =.458

concentation A- = 0.458 / 2.00 = .229 M

moles HA = 0.405 + 0.150 = 0.555

concentration HA = 0.555/ 2.00 L= 0.278 M

pH = 6.25 +log .229 / .278 = 6.165

Part C-

moles HA = 0.405 - 0.195 = 0.210

concentration HA = 0.210 / 2.00 = 0.105 M

moles A- = 0.608 + 0.195 =.803

concentration A- = 0.803 / 2.00= 0.4015 M

pH = 6.25 + log 0.4015 / 0.105 = 6.832