Answer:
a) 4.73
b) 4.78
c) 4.66 (further addition) or 4.60 (starting from the original buffer solution)
Explanation:
<u>Step 1:</u> Data given
volume of the buffer = 1.00 L
Buffer = 0.97 M CH3COONa / 1.02 M CH3COOH
pKa CH3COOH = 4.75
<u>Step 2: </u>pH = pKa + log [CH3COONa]/[CH3COOH]
pH = 4.75 + log (0.97/1.02)
pH =<u> 4.73</u>
(b) pH after addition of 0.065 mol NaOH
Adding 0.065 mol NaOH will reduce the acid by that amount leaving 1.02 - 0.065 = 0.955 moles HA in 1 L so [HA] = 0.955; the neutralized acid produces A- in the same amount, increasing [A-] to 0.97 +0.065 = 1.035
pH = pKa + log[CH3COONa]/[CH3COOH]
pH = 4.75 + log(1.035/0.955)
pH = <u>4.78</u>
c) pH after<u> further</u> addition of 0.144 mol HCl
The reverse will happen after the addition of HCl:
[HA] = 0.955 + 0.144 = 1.099
[A-] = 1.035 - 0.144 = 0.891
pH = 4.75 + log(0.891/1.099)
pH = 4.66
If we add 0.144 mol of HCl to the original buffer we will get:
[HA] = 1.02 + 0.144 = 1.164
[A-] = 0.97 - 0.144 = 0.826
pH = 4.75 + log(0.826/1.164)
pH = 4.60
