Question

What is the pH after 10.0 mL of 1.0 M NaOH have been added to 5.0 mL of 1.0 M HF?

Answer 1

The pH after adding 10.0 mL of 1.0 M NaOH to 5.0 mL of 1.0 M HF is 2.99.

Step-by-step explanation:

To calculate the pH after adding 10.0 mL of 1.0 M NaOH to 5.0 mL of 1.0 M HF, we need to determine the new concentrations of the buffer components.

First, we calculate the moles of NaOH and HF:

• Moles of NaOH = (10.0 mL)(1.0 M) = 10.0 mmol
• Moles of HF = (5.0 mL)(1.0 M) = 5.0 mmol

Then, we calculate the new volumes of NaOH and HF:

• New volume of NaOH = 10.0 mL + 5.0 mL = 15.0 mL
• New volume of HF = 5.0 mL

Next, we calculate the new concentrations of NaOH and HF:

• New concentration of NaOH = (10.0 mmol)/(15.0 mL) = 0.67 M
• New concentration of HF = (5.0 mmol)/(5.0 mL) = 1.0 M

Finally, we use the Henderson-Hasselbalch equation to calculate the pH:

pH = pKa + log([A-]/[HA])

Since HF is a weak acid and NaF is its conjugate base, we can use the pKa of HF (3.17) and the concentrations of NaF and HF:

• [A-] = 0.67 M
• [HA] = 1.0 M

Plugging these values into the equation gives us the pH:

pH = 3.17 + log(0.67/1.0) = 3.17 - 0.18 = 2.99