24 mL of 0.39 mol/L acetic acid is titrated with a standardized 0.33 mol/L KOH solution. Calculate the pH of the solution after 17 mL of the KOH solution has been added. Assume the Ka of acetic acid is 1.8 x 10-5.
To calculate the pH of the solution after 17 mL of the KOH solution has been added, we need to determine the concentration of the remaining acetic acid and the concentration of the resulting acetate ion.
First, let's calculate the number of moles of acetic acid initially present:
moles of acetic acid = volume of acetic acid (in L) * concentration of acetic acid (in mol/L)
moles of acetic acid = 0.024 L * 0.39 mol/L
moles of acetic acid = 0.00936 mol
Next, let's calculate the number of moles of KOH that reacted with the acetic acid:
moles of KOH = volume of KOH (in L) * concentration of KOH (in mol/L)
moles of KOH = 0.017 L * 0.33 mol/L
moles of KOH = 0.00561 mol
Since the reaction between acetic acid and KOH is 1:1, the number of moles of acetic acid remaining is the initial number of moles minus the moles of KOH reacted:
moles of acetic acid remaining = 0.00936 mol - 0.00561 mol
moles of acetic acid remaining = 0.00375 mol
Now, let's calculate the concentration of acetic acid remaining:
concentration of acetic acid = moles of acetic acid remaining / volume of solution remaining (in L)
volume of solution remaining = total volume of solution - volume of KOH added
volume of solution remaining = (0.024 L + 0.017 L) - 0.017 L
volume of solution remaining = 0.024 L
concentration of acetic acid remaining = 0.00375 mol / 0.024 L
concentration of acetic acid remaining = 0.15625 mol/L
Next, let's calculate the concentration of acetate ion:
Since the reaction between acetic acid and KOH is 1:1, the concentration of acetate ion is equal to the concentration of KOH used.
concentration of acetate ion = concentration of KOH used = 0.33 mol/L
Now, we can use the Henderson-Hasselbalch equation to calculate the pH:
pH = pKa + log10(acetate ion concentration / acetic acid concentration)
pKa = -log10(Ka)
pKa = -log10(1.8 x 10^-5)
pKa = 4.74
pH = 4.74 + log10(0.33 mol/L / 0.15625 mol/L)
pH = 4.74 + log10(2.108)
pH = 4.74 + 0.324
pH = 5.06
Therefore, the pH of the solution after 17 mL of the KOH solution has been added is approximately 5.06.
Write the acid/base equation.
Calculate mols KOH.
Calculate mols CH3COOH.
Calculate mols salt formed.
Calculate how much of the acid is unreacted.
The salt and acid form a buffered solutin. Use the Henderson-Hasselbalch equation to solve for pH.