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Multiple Choice
Calculate the pH of a solution made by adding 160 mL of 1.00 M HCl to 300 mL of 1.70 M NH3 solution.
A
pH = 9.25
B
pH = 8.95
C
pH = 4.75
D
pH = 7.00
Verified step by step guidance
1
Determine the moles of HCl added to the solution. Use the formula: \( \text{moles of HCl} = \text{volume (L)} \times \text{molarity (M)} \). Convert 160 mL to liters by dividing by 1000.
Calculate the moles of NH3 present in the solution using the formula: \( \text{moles of NH3} = \text{volume (L)} \times \text{molarity (M)} \). Convert 300 mL to liters by dividing by 1000.
Write the balanced chemical equation for the reaction between HCl and NH3: \( \text{HCl} + \text{NH}_3 \rightarrow \text{NH}_4^+ + \text{Cl}^- \). Determine the limiting reactant by comparing the moles of HCl and NH3.
Calculate the concentration of \( \text{NH}_4^+ \) ions formed in the solution. Use the formula: \( \text{concentration} = \frac{\text{moles of } \text{NH}_4^+}{\text{total volume of solution (L)}} \). Add the volumes of HCl and NH3 solutions to find the total volume.
Use the Henderson-Hasselbalch equation to find the pH of the solution: \( \text{pH} = \text{pK}_a + \log \left( \frac{[\text{base}]}{[\text{acid}]} \right) \). Identify \( \text{pK}_a \) for \( \text{NH}_4^+ \) and calculate the ratio of concentrations of \( \text{NH}_3 \) and \( \text{NH}_4^+ \).