Multiple Choice
What is the pH of a buffer solution after adding 0.150 mol of HCl, assuming no volume change?
17. Acid and Base Equilibrium
pH of Weak Acids
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What is the calculated pH after adding 2.5 mL of 0.5 M NaOH to 50.0 mL of 0.25 M acetic acid, assuming complete neutralization of the added base?
A
5.25
B
3.75
C
4.76
D
7.00
Verified step by step guidance1
Start by determining the moles of NaOH added. Use the formula: \( \text{moles} = \text{volume (L)} \times \text{molarity (M)} \). Convert 2.5 mL to liters by dividing by 1000, then multiply by the molarity of NaOH (0.5 M).
Calculate the initial moles of acetic acid present using the same formula: \( \text{moles} = \text{volume (L)} \times \text{molarity (M)} \). Convert 50.0 mL to liters and multiply by the molarity of acetic acid (0.25 M).
Determine the moles of acetic acid that react with NaOH. Since NaOH is a strong base, it will react completely with acetic acid. Subtract the moles of NaOH from the initial moles of acetic acid to find the moles of acetic acid remaining.
Calculate the moles of acetate ion formed, which will be equal to the moles of NaOH added, as each mole of NaOH reacts to form one mole of acetate ion.
Use the Henderson-Hasselbalch equation to find the pH: \( \text{pH} = \text{pK}_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \). The \( \text{pK}_a \) of acetic acid is approximately 4.76. Substitute the concentrations of acetate ion \([\text{A}^-]\) and acetic acid \([\text{HA}]\) into the equation, which are calculated by dividing the moles by the total volume of the solution (52.5 mL converted to liters).
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