The pH of a solution prepared by mixing 50.0 mL of 0.125 M KOH and 50.0 mL of 0.125 M HCl is __________.
i know the answer is 7.00, but im not sure how to solve it, do i create an ice chart?
To find the pH of the solution, you need to consider the reaction between KOH (a strong base) and HCl (a strong acid). In this reaction, they completely neutralize each other to form water (H2O) and potassium chloride (KCl).
The balanced chemical equation for the reaction is:
KOH + HCl → KCl + H2O
Since both KOH and HCl have the same molarity (0.125 M) and equal volumes (50.0 mL), you can assume that they will react in a 1:1 ratio. This means that all the KOH will react with all the HCl.
To solve the problem, you can follow these steps:
1. Calculate the number of moles of KOH:
0.125 M * 0.050 L = 0.00625 mol
2. Calculate the number of moles of HCl:
0.125 M * 0.050 L = 0.00625 mol
3. Since the reaction is 1:1, the limiting reagent is either KOH or HCl, whichever is less. In this case, both are equal, so either can be considered the limiting reagent. Since they are completely reacted, you will have no excess of either KOH or HCl.
4. Calculate the total volume of the solution:
50.0 mL + 50.0 mL = 100.0 mL = 0.100 L
5. Calculate the total moles of water formed (product) in the reaction:
0.00625 mol
6. Calculate the molarity of the water (product) in the solution:
0.00625 mol / 0.100 L = 0.0625 M
7. Water is a neutral substance, so the concentration of H+ and OH- ions are equal in pure water, which means the pH is 7.00.
Therefore, the pH of the solution prepared by mixing 50.0 mL of 0.125 M KOH and 50.0 mL of 0.125 M HCl is 7.00.