Question

I don't understand what I'm missing:

pH=-log(1.85x10^-5)+log(.49/.65)

= 4.69728

=4.70 (right?)

Part B What is the pH of a solution that is 0.49 M in sodium acetate and 0.65 M in acetic acid? (for acetic acid K, is 1.85x10.) IVO ACOM O a ? 4.71 O Check the rounding in your final answer. No credit lost. Try again. Submit Previous Answers Request Answer

Answer 1

Using Handerson Hasselbech equation; pH = pka + log [salt] NaoAC [cid] & ACOH = -log (1.85 x16 5) + log (oua ) = 4.7326 + (-0-1227) LPH = 4 | 082 . pH = M. | | Thur, pH of solution was found to be 4161