Multiple Choice
A 100.0 mL sample of 0.18 M HClO4 is titrated with 0.27 M LiOH. Determine the pH of the solution after the addition of 30.0 mL of LiOH.
18. Aqueous Equilibrium
Titrations: Strong Acid-Strong Base
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A volume of 10.0 mL of a 0.600 M HNO3 solution is titrated with 0.340 M KOH. Calculate the volume of KOH required to reach the equivalence point.
A
20.0 mL
B
10.0 mL
C
15.0 mL
D
17.6 mL
Verified step by step guidance1
Identify the balanced chemical equation for the reaction: \( \text{HNO}_3 + \text{KOH} \rightarrow \text{KNO}_3 + \text{H}_2\text{O} \). This shows a 1:1 molar ratio between HNO3 and KOH.
Calculate the moles of HNO3 present in the solution using the formula: \( \text{moles} = \text{concentration} \times \text{volume} \). Here, \( \text{concentration} = 0.600 \text{ M} \) and \( \text{volume} = 10.0 \text{ mL} \). Convert the volume to liters by dividing by 1000.
Since the reaction is a 1:1 molar ratio, the moles of KOH required will be equal to the moles of HNO3 calculated in the previous step.
Use the moles of KOH and its concentration to find the volume of KOH needed. Rearrange the formula \( \text{moles} = \text{concentration} \times \text{volume} \) to solve for volume: \( \text{volume} = \frac{\text{moles}}{\text{concentration}} \). Here, \( \text{concentration} = 0.340 \text{ M} \).
Convert the volume of KOH from liters to milliliters by multiplying by 1000 to find the final volume required to reach the equivalence point.
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