A buffer containing 0.30 M HCLO and 0.30 M NaClO has a pH of 7.53. What will the pH be after .010 mol of NaOH is added to 100 ml of the buffer.

Question

Answer options: 7.83
7.53
8.23
7.03

Answer 1

You want to use the HH equation (Henderson-Hasselbalch). The problem doesn't give you a pKa for HClO but you can calculate it.
pH = pKa + lob (base/acid)
7.53 = pKa + log (0.30/0.30)
pKa = 7.53
100 mL 0.30 M HClO = 30 millimoles HClO
100 mL 0.30 M NaClO = 30 millimiles NaClO
Add 0.010 mols NaOH = 10 millimoles NaOH added
..................HClO + OH^- ==> ClO^- + HOH
I...................30........0.................30................
add..........................10.....................................
C...............-10.........-10.............+10.......................
E..................20...........0.............+40....................
pH = pKa + log (base/acid)
pH = 7.53 + log (40/20). You can finish.
Technically, that 40/20 SHOULD BE (40/100/20/100). The 40/20 is the millimoles ratio. Technically concentration goes there which is 40/100 and 20/100 but I use millimoles almost exclusively since the volume ALWAYS WILL CANCEL. Post your work if you get stuck.
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