Multiple Choice
Calculate the volume (mL) of 0.500 M NaOH required to reach the equivalence point in the titration of 25.0 mL of 0.650 M HF. Ka for HF = 3.5 × 10−4.
18. Aqueous Equilibrium
Titrations: Weak Acid-Strong Base
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Determine the pH when 5.00 mL of excess KOH (beyond the equivalence point) has been added to the titration of 20.0 mL of 0.500 M HC2H3O2 with 0.350 M KOH.
A
1.49
B
2.75
C
12.07
D
12.32
E
12.51
Verified step by step guidance1
First, calculate the initial moles of acetic acid (HC2H3O2) using the formula: \( \text{moles} = \text{volume} \times \text{molarity} \). For HC2H3O2, use \( 20.0 \text{ mL} \) and \( 0.500 \text{ M} \).
Next, calculate the moles of KOH added using the same formula: \( \text{moles} = \text{volume} \times \text{molarity} \). For KOH, use \( 5.00 \text{ mL} \) and \( 0.350 \text{ M} \).
Determine the moles of KOH that reacted with HC2H3O2. Since KOH is added beyond the equivalence point, subtract the initial moles of HC2H3O2 from the moles of KOH to find the excess moles of KOH.
Calculate the concentration of excess OH\(^-\) ions in the solution. Use the formula: \( \text{concentration} = \frac{\text{moles of excess KOH}}{\text{total volume in liters}} \). The total volume is the sum of the volumes of HC2H3O2 and KOH.
Finally, determine the pH of the solution. Use the formula: \( \text{pH} = 14 - \text{pOH} \), where \( \text{pOH} = -\log[\text{OH}^-] \). Calculate \( \text{pOH} \) using the concentration of OH\(^-\) ions and then find the pH.
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