PART A: What is the pH of a buffer prepared by adding 0.809 mol of the weak acid
ID: 740116 • Letter: P
Question
PART A: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 2.00L 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.
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.
Explanation / Answer
[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 =5.82 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=5.46 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
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