Multiple Choice
A 25.00 mL sample of 0.175 M HCl is being titrated with 0.250 M NaOH. What is the pH after 17.50 mL of NaOH has been added?
18. Aqueous Equilibrium
Titrations: Strong Acid-Strong Base
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Consider the titration of 575.00 mL of HCl with a pH of 1.34. How much 0.0300 M NaOH (in mL) needs to be added to reach the equivalence point?
A
645.00 mL
B
690.00 mL
C
720.00 mL
D
575.00 mL
Verified step by step guidance1
Calculate the concentration of HCl using the pH value. The pH is given as 1.34, so use the formula \( \text{pH} = -\log[H^+] \) to find \([H^+]\).
Convert the concentration of \([H^+]\) to molarity. Since \([H^+] = 10^{-\text{pH}}\), substitute the pH value to find \([H^+]\) in mol/L.
Determine the moles of HCl present in the solution. Use the formula \( \text{moles} = \text{molarity} \times \text{volume (L)} \) to calculate the moles of HCl, where the volume is 575.00 mL converted to liters.
Set up the stoichiometry of the reaction between HCl and NaOH. The balanced chemical equation is \( \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \), which shows a 1:1 mole ratio.
Calculate the volume of 0.0300 M NaOH needed to reach the equivalence point. Use the formula \( \text{volume of NaOH (L)} = \frac{\text{moles of HCl}}{\text{molarity of NaOH}} \) and convert the volume from liters to milliliters.
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