Chemistry Forum

A laboratory technician uses a volumetric flask to prepare 500 mL of a 0.005 M solution of KOH (aq).
Subsequently, transfer 100.0 mL of the prepared solution to a flask. Forgetting to cap the bottle, the technician
Leave the solution in an open container for several days before use. During this time, part of the water evaporates and the volume
is reduced to 50.0 mL. What is the pH of the solution after evaporation, assuming a temperature of 25°C?

What are your own ideas.
Calculate the concentration in 100 ml and also in 50 ml.
pH = 14- pOH

I had started the account by calculating the initial number of moles and assuming that even transferring only 100 ml, the number of moles would remain constant, only the concentration and volume would change. Is that right?

so, initially we have: 0.0025 mol
with 100ml the concentration is 0.025
and with 50 ml the concentration is 0.05

I tried to find the PH starting from the remaining OH concentration of the base, but I couldn't calculate it. I assumed the reaction would be: KOH=K+OH

Your initial is correct 0,0025 mol, but then it goes wrong, if you take 100 ml then you have only 1/5 taken what gives 0,0005 mol

My calculation

First the Concentration is 0,005 M = 0,005 mol/l in 500 ml and also in 100 ml.
In 100 ml we have then 0,0005 mol = 0,5 mmol
The solution get concentrated by evaporation so the amount will be double reaching 50 ml . So the final ist 0,5 mmol/ 50 ml = 0,01 mol/l

The pOH is 2 what means the pH would be 12.

What is not considered is that the KOH will absorbance CO₂ from air and change to Carbonate., what gives lower pH.

I understood my mistake. So does that mean that whenever you take an aliquot of a solution, the number of moles will change and the concentration will remain constant? Besides that, as it evaporated, the number of moles remains constant, only what will change and the volume and concentration?

Yes think you want to check the salt content of a lake or river. You have millions gallons of water. You take only an aliquot und determine the concentration. Your result in mol/l or g/l in your sample is the same as in the big water.
The others you mentioned is right.

thank you so much, you are a genius