Question

Calculate the pH of 1.00 L of a buffer that is 0.100 M HNO₂ and 0.190 M NaNO₂. What is the pH of the same buffer after the addition of 1.00 mL of 12.0 M HCL. (pKa of HNO₂= 3.40) ( I need the Original buffer and and HCL after it's added)

Answer 1

The pH of the original buffer is 3.679. After adding 1.00 mL of 12 M HCl, the pH of the buffer changes to 3.601, demonstrating the buffer's ability to resist changes in pH.

Step-by-step explanation:

To calculate the pH of a buffer solution containing 0.100 M HNO2 and 0.190 M NaNO2, we use the Henderson-Hasselbalch equation:

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

Inserting the values we get:

pH = 3.40 + log(0.190/0.100)

pH = 3.40 + log(1.90)

pH = 3.40 + 0.279

pH = 3.679

After adding 1.00 mL of 12 M HCl to the buffer, the moles of HCl added are 0.012 moles. This will react with the NaNO2, reducing its concentration and increasing the concentration of HNO2.

New [NaNO2] = 0.190 - 0.012 = 0.178 M

New [HNO2] = 0.100 + 0.012 = 0.112 M

Now, we recalculate the pH with these new concentrations:

pH = 3.40 + log(0.178/0.112)

pH = 3.40 + log(1.589)

pH = 3.40 + 0.201

pH = 3.601

The original pH of the buffer is 3.679, and the pH after the addition of HCl is 3.601, showing the buffer's resistance to pH change.