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What is the pH of a solution prepared by dissolving 0.140 g KOH in sufficient water to prepare 250.0 ml of solution?
Okay to calculate this, you must first calculate the number of moles KOH forms in water. This is simply done by dividing mass by the molecular mass.
Therefore: 0.140 / (19+16+1)
Then, you need to know how KOH dissociates in water, which is luckily 1:1. Therefore, the number of moles of OH is also 0.140 / (19+16+1).
Then you find the concentration of OH, which is number of moles divided by the volume (do it in litres): (0.140 / (19+16+1))/0.25
Then, you find the pOH. This would be -log ((0.140 / (19+16+1))/0.25)
Then, you get this answer and minus if from 14: 14-(-log ((0.140 / (19+16+1))/0.25))
Make sure you actually do these steps as remember every problem is different so you need to know the process, not the answer.
The first step is to calculate the molarity of the KOH solution. Convert 0.140 g of KOH to moles and then determine the molarity. Since KOH is a strong base, it will fully dissociate in water to form KOH, resulting in a 1:1 ratio of KOH to OH-. Using the concentration of OH- ions, you can then convert it to pOH and then to pH. The pH of the solution will be basic due to the presence of KOH.
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