Questions: Question 7 (1 point) A student is making a titration curve. To do so, they added 25.00 mL of 0.2470 M HCl to a flask. They then added 49.14 mL of 0.1170 M NaOH to the flask. What is the pH at this point in the titration curve? 3.1117 1.9343 2.2406 3.3436 1.6537

Solution Steps

First, calculate the moles of HCl and NaOH added to the flask.

• Moles of HCl: $$\text{Moles of HCl} = 25.00 \, \text{mL} \times 0.2470 \, \text{M} = 0.02500 \, \text{L} \times 0.2470 \, \text{mol/L} = 0.006175 \, \text{mol}$$

• Moles of NaOH: $$\text{Moles of NaOH} = 49.14 \, \text{mL} \times 0.1170 \, \text{M} = 0.04914 \, \text{L} \times 0.1170 \, \text{mol/L} = 0.005750 \, \text{mol}$$

Since HCl and NaOH react in a 1:1 ratio, determine the limiting reactant and the remaining moles of the excess reactant.

• Since 0.006175 mol of HCl is greater than 0.005750 mol of NaOH, NaOH is the limiting reactant.

• Remaining moles of HCl after reaction: $$\text{Remaining moles of HCl} = 0.006175 \, \text{mol} - 0.005750 \, \text{mol} = 0.000425 \, \text{mol}$$

Calculate the concentration of the remaining HCl in the total volume of the solution to find the pH.

• Total volume of the solution: $$\text{Total volume} = 25.00 \, \text{mL} + 49.14 \, \text{mL} = 74.14 \, \text{mL} = 0.07414 \, \text{L}$$

• Concentration of remaining HCl: $$[\text{H}^+] = \frac{0.000425 \, \text{mol}}{0.07414 \, \text{L}} = 0.005731 \, \text{M}$$

• pH calculation: $$\text{pH} = -\log_{10}(0.005731) \approx 2.2406$$

Final Answer