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 and the pH of the buffer after the addition of NaOH, we need to first understand the composition of the buffer and then use the Henderson-Hasselbalch equation.

A buffer is a solution that resists changes in pH when small amounts of acid or base are added. It consists of a weak acid and its conjugate base or a weak base and its conjugate acid. In this case, the buffer is made by combining butylamine (C4H9NH2) and butylammonium chloride (C4H9NH3Cl).

A) To calculate the pH of the buffer, we can use the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])

In this case, butylamine (C4H9NH2) acts as a weak base (B) and butylammonium chloride (C4H9NH3Cl) acts as a conjugate acid (BH+).

The pKa value can be found from the dissociation constant of butylammonium chloride (C4H9NH3Cl), which is the conjugate acid:
C4H9NH3Cl ⇌ C4H9NH2 + H+

The pKa value for butylammonium chloride can be obtained from reliable sources or calculated experimentally. Let's assume it is 4.75 for this example.

Now, we can substitute the given values into the Henderson-Hasselbalch equation:
pH = 4.75 + log ([C4H9NH2]/[C4H9NH3Cl])

To calculate the concentrations of butylamine and butylammonium chloride, we can use the given molarity (M) and volume (L) information. Remember, volumes are additive, so the total volume of the buffer solution is the sum of the volumes of the two initial solutions.

Volume of butylamine: 3.50 L
Molarity of butylamine: 0.200 M
Concentration of butylamine ([C4H9NH2]) = Molarity × Volume = 0.200 M × 3.50 L = 0.70 mol

Volume of butylammonium chloride: 7.50 L
Molarity of butylammonium chloride: 0.100 M
Concentration of butylammonium chloride ([C4H9NH3Cl]) = Molarity × Volume = 0.100 M × 7.50 L = 0.75 mol

Now, substitute these concentrations into the Henderson-Hasselbalch equation:
pH = 4.75 + log (0.70/0.75)
Simplify the logarithm calculation:
pH = 4.75 + log (0.93)

Use a scientific calculator to calculate the log value and then add it to 4.75:
pH ≈ 4.75 + (-0.030)

The pH of the buffer solution is approximately 4.72.

B) To calculate the pH of the buffer after the addition of NaOH, we need to consider the reaction between the added NaOH and butylamine (C4H9NH2).

From the balanced equation:
NaOH + C4H9NH2 ⇌ C4H9NH3OH + Na+

The added NaOH reacts with butylamine, converting it into butylamine-hydroxide (C4H9NH3OH).

Given:
Volume of NaOH: 35.0 mL = 0.035 L
Molarity of NaOH: 0.080 M

Calculate the number of moles of NaOH:
Number of moles of NaOH = Molarity × Volume = 0.080 M × 0.035 L = 0.0028 mol

As we know, butylamine reacts with NaOH in a 1:1 ratio. Therefore, 0.0028 mol of butylamine will react with 0.0028 mol of NaOH.

Now, we need to determine how much butylamine and butylammonium chloride are remaining after the reaction. Subtract the moles of NaOH reacted from the initial moles of butylamine and butylammonium chloride.

Moles of butylamine remaining = Initial moles of butylamine - Moles of NaOH reacted = 0.70 mol - 0.0028 mol = 0.6972 mol

Moles of butylammonium chloride remaining = Initial moles of butylammonium chloride = 0.75 mol (as it does not react)

Now, we can use the Henderson-Hasselbalch equation again to calculate the new pH of the buffer after the addition of NaOH.

pH = 4.75 + log ([C4H9NH2]/[C4H9NH3Cl])
pH = 4.75 + log (0.6972/0.75)

Simplify and calculate the logarithm:
pH ≈ 4.75 + (-0.036)

The pH of the buffer solution after the addition of 35.0 mL of 0.080 M NaOH is approximately 4.71.