K benzoic acid = 6.6 x 10-5
PH = PKa + log (A/HA)
PH = -log(6.6x10^-5) + log(.210 M / .130 M)
PH = 4.18 + .21 = 4.39
Content will be erased after question is completed.
Enter the email address associated with your account, and we will email you a link to reset your password.
Forgot your password?