When studying chemistry, one topic you will surely come upon in your coursework is acid-base chemistry. Although the detail in which you'll delve into this area probably depends upon your chosen academic discipline, you will almost certainly study the concept of pH, a value which measures a liquid's acidity or alkalinity. When dealing with acids, another important value is pKa, which represents how readily the acid dissociates in solution. If you know an acid's concentration and pKa value, you can calculate its pH.

Write the equation for Ka: Ka = [H+][A-]/[HA]. In this equation, Ka is the negative, inverse logarithm of pKa -- in other words, pKa = -log(Ka) -- [H+] is the concentration of hydrogen ions in the solution, [A-] is the concentration of conjugate base and [HA] is the concentration of the acid. For the sake of example, suppose you're dealing with an acid with a concentration of 0.02 M.

Keep in mind that both the [H+] and [A-] values for an acid already in solution will be zero at the time of dissociation and "x" once dissociation is complete -- and that [HA] will decrease by "x" over this same time frame. Rewrite the equation algebraically: Ka = (x)(x)/(0.02 - x) = x^2/(0.02 - x). You can further modify it to form a quadratic equation: (Ka)(0.02 - x) = x^2/(0.02 - x), or 0.02KA - Kax = x^2. Further simplify the expression by getting everything on one side, as follows: 0.02Ka - Kax - .02Ka + Kax = x^2 - .02Ka + Kax. Therefore, x^2 - .02Ka + Kax = 0.

Solve this equation for "x" (in other words, [H+]), keeping in mind that the generic solution for any quadratic equation ax^2 + bx + c = 0 is x = [-b + sqrt(b^2 - 4ac)]/2a. For your example equation, then, this would be x = [-Ka + sqrt(Ka^2+0.08_Ka)]/2. Convert your "pKa" -- in other words, the negative logarithm of Ka -- into Ka by entering it into your scientific calculator, pressing the negative key and then the "10^x" key. If, for example, your pKa value is 5.4, your Ka would be 10^-5.4, or .00000398. Plug this value into your equation to solve, which for the example would be as follows: x = [-0.00000398 + sqrt(0.00000398^2+0.08_0.00000398)]/2 = [-0.00000398 + sqrt(0.0000000000158+0.0000003184)]/2 = [-0.00000398 + sqrt(0.0000003184158)]/2 = [-0.00000398 + 0.000564]/2 = 0.000560/2 = 0.000280.

Take the logarithm of your "x" -- again, your [H+] value -- and make it negative to calculate pH. For the example acid with a concentration of 0.02 M and pKa value of 5.4, this would be log[0.000280] = -3.55, made negative to give a pH of 3.55.