1)Calculate the pH of the solution after the addition of 8.58×10^{-2} mol of NaOH(s).

Please note that, here we have:

[H+] = (8580-8.58v)/(1000+v) * 10^(-2), where v is the volume of NaOH added

so using the formula for the Equilibrium constant, we get:

v = 149 mL

and:

[H+] = 0.06354

so:

pH = 1.197

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