How many moles of sodium hypobromite (NaBrO) should be added to 1.00 L of 5.5×10^?2 M hypobromous acid (HBrO) to form a buffer solution of pH= 9.17? Assume that no volume change occurs when the NaBrO is added. I posted this question previously but it was answered incorrectly
Homework Answers
Formula of buffer for weak acid and strong base is
pH= pKa + log[salt]/[acid]
pKa for hypobromous acid= 8.65 since its Ka value=2.8×10-9
Now, putting the values pH=9.17, pka=8.65 , [HBrO]=5.5×10-2M
9.17=8.65+log[NaBrO]/[HBrO]
0.52=log[NaBrO]/[HBrO]
3.311=[NaBrO]/[HBrO]
[NaBrO]=0.1821 moles
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