Final answer:
To calculate the pH after adding 5.51 mL of 0.150 mol L⁻¹ NaOH to a 12.7 mL 0.100 mol L⁻¹ HF solution, the remaining moles of HF is determined after the reaction with NaOH, and then the pH is calculated from the new H⁺ concentration, resulting in a pH of approximately 1.66.
Step-by-step explanation:
A student asked the question about calculating the pH after adding 5.51 mL of 0.150 mol L⁻¹ NaOH to 12.7 mL of a 0.100 mol L⁻¹ HF solution in a titration process.
In order to solve this, we must first find out how many moles of NaOH have been added:
5.51 mL × 0.150 mol/L = 0.0008265 mol NaOH
The number of moles of HF at the beginning would be:
12.7 mL × 0.100 mol/L = 0.00127 mol HF
Next, because NaOH is a strong base and HF a weak acid, they will react in a 1:1 molar ratio:
0.00127 mol HF - 0.0008265 mol NaOH = 0.0004435 mol HF remaining
The concentration of remaining HF is then:
0.0004435 mol / (12.7 mL + 5.51 mL) × 1000 mL/L = 0.02205 mol/L
The dissociation of HF can be represented as:
HF (aq) → H⁺ (aq) + F⁻ (aq)
Since we know HF is a weak acid, we can assume the concentration of H⁺ ions is approximately equal to the molarity of HF because the degree of dissociation is low. The pH can therefore be approximated by taking the negative logarithm of the H⁺ ion concentration:
- pH = -log([H⁺])
- pH = -log(0.02205)
- pH = 1.66
Given the information, the pH after adding 5.51 mL of NaOH is approximately 1.66.