How would you find the pH of a solution prepared by mixing 50 mL of 0.125 M KOH with o.050 L of 0.125 M HCl?
mols KOH = M x L = ?
mols HCl = M x L = ?
Determine which is in excess, the
pH = -log(H^+) if HCl is in excess
pOH = -log(OH^-) if KOH is in excess.
pH = 7 if neither is in excess.
To find the pH of the solution prepared by mixing KOH (a base) with HCl (an acid), you need to consider the chemical reaction that occurs between the two. KOH and HCl react in a 1:1 ratio, forming water (H2O) and a salt, potassium chloride (KCl).
Let's calculate the moles of KOH and HCl to understand the chemical reaction:
Moles of KOH = concentration (M) x volume (L) = 0.125 M x 0.050 L = 0.00625 moles
Moles of HCl = concentration (M) x volume (L) = 0.125 M x 0.050 L = 0.00625 moles
Since the reaction occurs in a 1:1 ratio, the moles of KOH and HCl are equal.
Now, let's find the total volume of the solution:
Total volume = volume of KOH + volume of HCl = 50 mL + 0.050 L = 0.050 L + 0.050 L = 0.100 L
To calculate the concentration of the final solution, divide the moles by the total volume:
Concentration = moles / volume = 0.00625 moles / 0.100 L = 0.0625 M
Now that we have the concentration of the final solution, we can calculate the pOH and pH. The pOH is obtained by taking the negative logarithm base 10 of the hydroxide concentration ([OH-]), and the pH is obtained by subtracting the pOH from 14:
pOH = -log[OH-]
pH = 14 - pOH
Since KOH is a strong base and HCl is a strong acid, they react completely, resulting in the formation of water. In water, the concentration of hydroxide ions (OH-) and hydrogen ions (H+) are equal:
[OH-] = [H+]
Therefore, the pOH of the solution is equal to the pH. To find the pH, calculate the negative logarithm base 10 of the concentration:
pH = -log[H+] = -log(0.0625) = 1.2
Thus, the pH of the solution prepared by mixing 50 mL of 0.125 M KOH with 0.050 L of 0.125 M HCl is 1.2.