Question

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

Answer 1

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