pH of a 1.9 × 10⁻³ M RbOH solution

Question

What is the pH of a 1.9 × 10⁻³ M RbOH solution?

I used -log, and the answer I got was ~2.72

The software thing we use is saying I calculated the pOH, so what's the difference, and how do I solve for just pH?

Answer 1

As RbOH dissociates into Rb¹⁺ and OH^1-, the OH^1- concentration is also 1.9 x 10^-3. [H¹⁺] [OH^1-] = 10^-14, so

the [H¹⁺] = 10^-14 / 1.9 x 10^-3 = 5.3 x 10^-12 . Take the negative log of this to get pH = 11.3.

The log of 1.9 x 10^-3 is -2.72, but pOH is the negative of this which is 2.72.

pH + pOH = 14, so you can calculate pH from this also.

Answer 2

pH is the is the power/potential of hydrogen ion. Its something that measures how much H⁺ or H₃O⁺ there is in a solution aka the acidity. pOH is the exact opposite, it is the power/potential of hydroxide ion, OH^-. Aka how basic the solution is. We very often use pH to standardize the measure of acid and measure of base. So on a pH test strip, 1 is super acid and 14 is super base. but on a pOH scale, 1 is super base and 14 is super acid. This is mainly because they are inversely proportional to each other. In STP, they are related to each other using the formula 14 = pH + pOH, or if you want it in terms of concentration, 1x10^-14=[H⁺] x [OH^-].

Going back to your question. You did the -log for the concentration of OH in RbOH. That is why you got pOH. You can use pOH to solve for pH when it is at STP. Using 14 = pH + pOH, you can isolate pH. So, 14= pH + 2.72 ---> pH=11.27.

In the future, you can also use 1x10^-14=[H⁺] x [OH^-] if the question allows.