Multiple Choice
Two 20.0-mL samples, one 0.200 M KOH and the other 0.200 M unknown base, are titrated with 0.100 M HI. What is the volume of added acid at the equivalence point for each titration?
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18. Aqueous Equilibrium
Titrations: Strong Acid-Strong Base
Multiple Choice
Calculate the pH of the solution resulting from the mixing of 175.0 mL of 0.250 M HNO3 with 75.0 mL of 0.200M Ba(OH)2.
A
pH = 12.74
B
pH = 0.94
C
pH = 1.26
D
pH = 13.06
Verified step by step guidance1
Determine the moles of HNO3 by using the formula: \( \text{moles of HNO3} = \text{volume (L)} \times \text{molarity (M)} \). Convert 175.0 mL to liters and multiply by 0.250 M.
Determine the moles of Ba(OH)2 using the formula: \( \text{moles of Ba(OH)2} = \text{volume (L)} \times \text{molarity (M)} \). Convert 75.0 mL to liters and multiply by 0.200 M.
Recognize that Ba(OH)2 is a strong base that dissociates completely in water to produce 2 moles of OH⁻ ions per mole of Ba(OH)2. Calculate the total moles of OH⁻ ions produced.
Calculate the moles of H⁺ ions remaining after the neutralization reaction between HNO3 and OH⁻ ions. Use the stoichiometry of the reaction: \( \text{HNO3} + \text{OH}^- \rightarrow \text{H2O} + \text{NO3}^- \). Subtract the moles of OH⁻ from the moles of HNO3.
Calculate the pH of the resulting solution using the formula: \( \text{pH} = -\log[\text{H}^+] \), where \([\text{H}^+]\) is the concentration of H⁺ ions remaining in the solution. Divide the moles of H⁺ by the total volume of the solution in liters to find the concentration.
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