Potassium hydroxide, KOH is a strong base and show great pH values in aqueous solution. In this tutorial, we are going to learn how to calculate pH of different concentration KOH aqueous solutions.
At the end of this tutorial, there is an online calculator which can be used to check your answers (pH values) after you finish your problems.
written by: Heshan Nipuna, last update: 03/06/2020
When calculating pH of KOH solutions, following theories and equations are important.
Question #1
In the laboratory, there is a aqueous potassium hydroxide solution which was prepared by a student. In the label of the bottle, concentration was mentioned as 0.1 mol dm-3. Calculate the theoretical pH value of the KOH solution.
Answer
According to the theory 1, OH- concentration is 0.1 mol dm-3
Now, you can use the equation 3 to calculate pOH with known OH- concentration.
pOH = -log10[OH-(aq)]
pOH = -log10[0.1]
pOH = 1
Now, use the equation 3 to find the pH
pH + pOH = 14
pH + 1 = 14
pH = 13
pH of 0.1 mol dm-3 KOH solution is 13.
Question #2
As a student you are curious to know about chemicals and their application in the laboratory. You found the analytical grade potassium hydroxide bottle in the laboratory. In the bottle, there are potassium hydroxide pellets. You got some of these pellets and measured the mass. In the balance mass of KOH is shown as 1.4 g. Now one of your friend came to the lab and ask about what's going on. You explained the stuff you had done and your friend want to continue the work. So he fill distilled water to a flask upto 250 cm3 and say you to put your measured KOH to it. Then you put them to water and mix the solution very well. You say KOH solution should be a basic one and should have a pH value greater than 7. So your friend is not sure about that and ask to prove it. Both of you go to the pH meter and check pH of the solution and it confirms you are correct. What will the pH of the solution?
K = 39 , O = 16 , H = 1
Answer: pH = 13
In this calculator, we use several methods to calculate pH of KOH solution. You can use one of it according to the data you have.
In our calculators pH of the neutral solution is 7 according to the 250C.
Method 1
Answer
Method 2
We assume that after adding KOH, volume of solute is not changed much. So you can enter volume of solution as volume of solute.
Answer
Method 3
We assume that after adding KOH, volume of solute is not changed much. So you can enter volume of solution as volume of solute.
Answer
We can take several KOH solutions and measure their pH values. Here are some KOH solutions with their concentration (mol dm-3).
When concentration of KOH decreases, pH also decreases.
Questions
Ask your question and find the answer free.First step is calculate the pOH of KOH solution. pOH of 1.0 * 10-2 mol-1 is 2. Then, calculate the pH by using relation of pH + pOH = 14. So pH of the 1.0 * 10-2 mol-1 KOH is 12.
Because you know the pH value, you can find the pOH. pOH = 14 - pH . pOH of KOH solution is 1.36
Now use the pOH equation.
OH- ion is provided by only KOH. Therefore concentration of KOH = 0.044 mol dm-3
If concentration of KOH is 0.1 M, pH value may be 13. If concentation is 0.0001 M, pH value will be 10. Therefore, pH range of KOH solution will be 10-13.
At most cases, this can be true because potassium hydroxide is a strong base. Ammonia give very less OH- ions in aqueous state.
Both Potassium hydroxide and Barium hydroxide are strong bases. Though both have same concentration, their OH- concentration is not same.
Barium hydroxide is a strong base and dissociate like below.
Ba(OH)2 → Ba2+ + 2OH-
OH- concentration of Barium hydroxide is greater than OH- concentration of KOH. Therefore pH of Barium hydroxide solution is greater than KOH solution.
OH- concentration of Ba(OH)2 solution is greater because when one mol of KOH dissociates, two moles of OH- are given. But, for KOH, one mol of KOH only give one mol of OH-.
When potassium metal is added to water, a strong basic solution, KOH is formed. Due to this strong basic solution, pOH value will be low and pH value will be high.