I am stuck in the PLUG-N-CHUG PART. Here is what I get when I plug-n-chug...1.75 x 10-5 + log[0.164M/0.225M] I get the pH as 4.62 not confident in answer....please guide me to correct answer
Check to make sure I've substituted correctly since the original numbers are back at the other post. From what you posted, did you substitute 1.75E-5 for Ka. That's Ka. Pka = 4.76.
pH = pKa + log (0.164/0.225)
pH = 4.76 + log0.729
pH = 4.76 - 0.137 = 4.46