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QUESTION

# How do you calculate pH from acid dissociation constant?

For a weak acid, set up the equilibrium expression for dissociation to ions in solution, then solve this equation for the hydronium ion concentration. The can be calculated directly from [H^+].

Example: The pH of 0.2 M acetic acid (HOAc)

HOAC &harr; H^+ + OAc^- K_a = 1.8x10^(-5) = ([H^+][OAc^-])/([HOAC]) If the acid is weak, then only a small concentration, x, will dissociate. We can rewrite the equation as K_a = 1.8x10^(-5) = ([H^+][OAc^-])/([HOAC]) = (x^2)/(0.2-x) & approximately; x^2/0.2where we have ignored x in the denominator because it is a small number compared with 0.2. Solving this equation gives * See note below about what to do if you do not want to ignore this... x = (0.2 &middot; 1.8x10^(-5))^(1/2) = 1.9x10^(-3) = [H^+] pH = -log([H^+]) = 2.72

NOTE: If you don't simpify 0.2-x = 0.2, you need to use the quadratic equation to solve for the pH. Here is a video which discusses how to do this.