2.Part A What is the pH of a buffer prepared by adding 0.708 m o l of the weak a
ID: 798434 • Letter: 2
Question
2.Part A What is the pH of a buffer prepared by adding 0.708mol of the weak acid HA to 0.609mol of NaA in 2.00 L of solution? The dissociation constant Ka of HA is 5.6610?7. Ph= 6.18 Part B What is the pH after 0.150mol of HCl is added to the buffer from Part A? Assume no volume change on the addition of the acid. Part C What is the pH after 0.195mol of NaOH is added to the buffer from Part A? Assume no volume change on the addition of the base.
2.Part A What is the pH of a buffer prepared by adding 0.708mol of the weak acid HA to 0.609mol of NaA in 2.00 L of solution? The dissociation constant Ka of HA is 5.6610?7. Ph= 6.18 Part B What is the pH after 0.150mol of HCl is added to the buffer from Part A? Assume no volume change on the addition of the acid. Part C What is the pH after 0.195mol of NaOH is added to the buffer from Part A? Assume no volume change on the addition of the base.
2.Part A What is the pH of a buffer prepared by adding 0.708mol of the weak acid HA to 0.609mol of NaA in 2.00 L of solution? The dissociation constant Ka of HA is 5.6610?7. Ph= 6.18 Part B What is the pH after 0.150mol of HCl is added to the buffer from Part A? Assume no volume change on the addition of the acid. Part C What is the pH after 0.195mol of NaOH is added to the buffer from Part A? Assume no volume change on the addition of the base.
2.Part A What is the pH of a buffer prepared by adding 0.708mol of the weak acid HA to 0.609mol of NaA in 2.00 L of solution? The dissociation constant Ka of HA is 5.6610?7. Ph= 6.18 Part B What is the pH after 0.150mol of HCl is added to the buffer from Part A? Assume no volume change on the addition of the acid. Part C What is the pH after 0.195mol of NaOH is added to the buffer from Part A? Assume no volume change on the addition of the base.
Explanation / Answer
[HA] = 0.708/ 2.00 = 0.354 M
[A-] = 0.609/2.00 =0.305 M
pKa = 6.25
pH = 6.25 + log 0.305 / 0.354 =6.19
A- + H+ >> HA
moles A- = 0.609 - 0.150 =0.459
[A-]= 0.459/ 2.00 =0.230 M
moles HA = 0.708 + 0.150 =0.858
[HA] = 0.858/2.00 = 0.429 M
pH = 6.25 + log 0.230/0.429 = 5.98
HA + OH- >> A- + H2O
moles HA = 0.708 - 0.195 =0.513
[HA]= 0.513/2.00 =0.257 M
moles A- = 0.609 + 0.195 =0.804
[A-]= 0.804/2.00 = 0.402 M
pH = 6.25 + log 0.402/0.257 =6.44
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