A buffer is made by combining 3.50 L of 0.200M butylamine, C4H9NH2 with 7.50L of 0.100M butylammonium chloride, C4H9NH3Cl. Assuming that volumes are additive, calculate the following.

A) the pH of the buffer. B) the pH of the buffer after the addition of 35.0mL of 0.080M NaOH to the 525.mL of the buffer.

To calculate the pH of the buffer, we need to consider the dissociation of the butylamine and the butylammonium chloride in water.

Step 1: Calculate the moles of butylamine and butylammonium chloride:
- Moles of butylamine = volume (L) × molarity (M) = 3.50 L × 0.200 M = 0.70 moles
- Moles of butylammonium chloride = volume (L) × molarity (M) = 7.50 L × 0.100 M = 0.75 moles

Step 2: Calculate the total moles of the buffer:
- Total moles = moles of butylamine + moles of butylammonium chloride = 0.70 moles + 0.75 moles = 1.45 moles

Step 3: Calculate the concentration of the buffer:
- Concentration of the buffer = total moles / total volume (L) = 1.45 moles / (3.50 L + 7.50 L) = 0.145 M

Step 4: Calculate the pOH of the buffer:
- pOH = -log10 (concentration of the buffer)
- pOH = -log10 (0.145 M) ≈ 0.839

Step 5: Calculate the pH of the buffer:
- pH + pOH = 14 (since we are in neutral conditions)
- pH = 14 - pOH = 14 - 0.839 ≈ 13.161

So, the pH of the buffer is approximately 13.161.

Now, let's calculate the pH of the buffer after the addition of 35.0 mL of 0.080 M NaOH to the 525. mL of the buffer.

Step 1: Calculate the moles of NaOH added:
- Moles of NaOH = volume (L) × molarity (M) = 0.035 L × 0.080 M = 0.0028 moles

Step 2: Calculate the moles of butylamine and butylammonium chloride remaining in the buffer:
- Moles of butylamine remaining = initial moles of butylamine - moles of NaOH added = 0.70 moles - 0.0028 moles = 0.6972 moles
- Moles of butylammonium chloride remaining = initial moles of butylammonium chloride - moles of NaOH added = 0.75 moles - 0.0028 moles = 0.7472 moles

Step 3: Calculate the new total moles of the buffer:
- New total moles = moles of butylamine remaining + moles of butylammonium chloride remaining = 0.6972 moles + 0.7472 moles = 1.4444 moles

Step 4: Calculate the new concentration of the buffer:
- New concentration of the buffer = new total moles / total volume (L) = 1.4444 moles / (0.525 L) = 2.75 M

Step 5: Calculate the new pOH of the buffer:
- new pOH = -log10 (new concentration of the buffer)
- new pOH = -log10 (2.75 M) ≈ -0.438

Step 6: Calculate the new pH of the buffer:
- new pH + new pOH = 14 (since we are in neutral conditions)
- new pH = 14 - new pOH = 14 - (-0.438) ≈ 14.438

So, the pH of the buffer after the addition of 35.0 mL of 0.080 M NaOH to the 525. mL of the buffer is approximately 14.438.