Multiple Choice
What is the pH at the equivalence point in the titration of 100 mL of 0.10 M HCl with 0.10 M NaOH?
10. Acid-Base Titrations
Strong Acid-Strong Base Titrations
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During the titration of 25.0 mL of 0.100 M HCl with 0.100 M NaOH, what is the pH after adding 15.0 mL of NaOH?
A
12.70
B
1.70
C
2.00
D
1.30
Verified step by step guidance1
Determine the moles of HCl initially present using the formula: \( \text{moles of HCl} = M_{\text{HCl}} \times V_{\text{HCl}} \), where \( M_{\text{HCl}} = 0.100 \, \text{M} \) and \( V_{\text{HCl}} = 25.0 \, \text{mL} \) (convert to liters).
Calculate the moles of NaOH added using the formula: \( \text{moles of NaOH} = M_{\text{NaOH}} \times V_{\text{NaOH}} \), where \( M_{\text{NaOH}} = 0.100 \, \text{M} \) and \( V_{\text{NaOH}} = 15.0 \, \text{mL} \) (convert to liters).
Determine the limiting reactant by comparing the moles of HCl and NaOH. Subtract the smaller value from the larger to find the moles of the excess reactant remaining after the reaction.
If HCl is in excess, calculate the concentration of \( \text{H}^+ \) ions using \( [\text{H}^+] = \frac{\text{moles of excess HCl}}{\text{total volume of solution in liters}} \). Then, find the pH using \( \text{pH} = -\log_{10}([\text{H}^+]) \).
If NaOH is in excess, calculate the concentration of \( \text{OH}^- \) ions using \( [\text{OH}^-] = \frac{\text{moles of excess NaOH}}{\text{total volume of solution in liters}} \). Then, find the pOH using \( \text{pOH} = -\log_{10}([\text{OH}^-]) \), and convert to pH using \( \text{pH} = 14 - \text{pOH} \).
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