The pH of a urine sample was measure to be 5.53. What is the [H₃O⁺] and [OH⁻] and pOH of the urine.

Answer 1

First, we have to remember what pH and pOH are. pH is calculated by getting -log[H+] while pOH is calculated by getting -log[OH-]. This question is somewhat of a reverse of this as you are already given the pH.

Second, we need to determine the pOH of the solution. It is important to remember that pH + pOH = 14. Therefore, if your pH is 5.53, the pOH must be 14 - 5.53 which is 8.47.

To finally answer the question, we need to think about how to get rid of our -log in both the pH and pOH equations. To remove the logarithm, we would make it have a base of 10, making our -log[H+] or [OH-] an exponent such as 10 ^ (-log[H+]) or 10 ^ (-pH). Now the logarithm is cancelled, we would keep the negative sign in front of the number, denoting that our answer is in decimals. This makes sense because the concentrations of both our H+ and OH- should add up to 1.

Therefore, the pH of the solution is 10 ^ -5.53 and the pOH is 10 ^ -8.47.