What is the pH at the equivalence point in the titration of 20.0 mL of 0.200 M HA with 0.200 M NaOH? The Ka for HA is 7.07×10−6 .

here the number of equivalen of the acid is:

4*10^(-3), so we have to require that the base has 4*10^(-3) number of equivalent, so the volume of the added base is:

(0.2/1000)*V = 4*10^(-3)

and then:

V = 20 mL

so the concentration of the resultant salt Cs is:

Cs= (4*10^(-3))/(40*10^(-3)) = 0.2 N

then the [OH-] concentration of our solution is:

[OH-] = sqrt(Kw Cs/Ka) = sqrt(10^(-14)*0.1/ (7.07*10^(-6)) = 1.19*10^(-5)

so the pOH = 4.92 and pH = 14- 4.92 = 9.08

please note that Kw is the ionic product of the water

Secure Information

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?

Sign Up