The pH of a 250.mL sample of a buffer solution is 9.85.If 1.0 mL of 6 M HCl is added,the pH of the resulting mixture is closest to
A)0.00.
B)1.65.
C)7.00.
D)9.70.
E)10.00.

Question

Quizplus · Accepted Answer

First, we need to determine the components of the buffer solution and its pH. Since the pH is greater than 7, we know that the buffer must contain a weak base and its conjugate acid. Let's assume this is ammonia (NH3) and ammonium (NH4+), respectively:

NH3 + H2O ↔ NH4+ + OH-

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

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

where pKa is the dissociation constant of the weak acid (in this case, NH4+), [A-] is the concentration of the weak base (NH3), and [HA] is the concentration of the conjugate acid (NH4+). We don't know the exact concentrations of these species, but we can assume they are roughly equal (i.e. the buffer is at or near the midpoint of its buffering range), so we can simplify the equation to:

pH = pKa + log(1) = pKa

We can look up the pKa of ammonium in a table (9.25), so:

pH = 9.25

Now, we add 1 mL of 6 M HCl to the buffer. The H+ ions react with the OH- ions in the buffer:

HCl + OH- → H2O + Cl-

The amount of OH- present in the buffer is determined by the equilibrium between ammonia and ammonium. When we add H+ ions, the equilibrium shifts to the left to consume some of the added H+ ions, which decreases the concentration of OH- and increases the concentration of NH4+:

NH3 + H+ ↔ NH4+

The pH of the resulting mixture will depend on how much the equilibrium shifts. We can use the Henderson-Hasselbalch equation again, but this time with the new concentration of NH4+ and NH3. Let's assume that all of the added H+ ions are consumed by the buffer (this is not necessarily true, but it makes the calculation simpler):

moles of H+ = 1 mL x (6 mol/L) = 0.006 mol

The buffer contains 250 mL x (0.1 mol/L) = 0.025 mol of NH3 and NH4+ total, so the initial concentration of each is:

[NH3] = [NH4+] = 0.025 mol / 0.25 L = 0.1 mol/L

After adding H+ ions, the concentration of NH4+ has increased by 0.006 mol/L, so:

[NH4+] = 0.1 + 0.006 = 0.106 mol/L

The concentration of NH3 will have decreased by the same amount, but we can't simply subtract since the initial buffer contains some NH3 and NH4+ already in equilibrium. Instead, we need to use the equilibrium constant (Ka) to calculate the new concentration of NH3:

Ka = [NH4+][OH-] / [NH3]

We know that [OH-] = 10^-4.15 M (since the pH of the buffer is 9.85, so pOH = 4.15), and we can solve for [NH3]:

Ka = 5.6 x 10^-10 (for NH4+)
[NH3] = [NH4+][OH-] / Ka = (0.106)(10^-4.15) / (5.6 x 10^-10) = 2.0 x 10^-4 M

Now we can use the Henderson-Hasselbalch equation again:

pH = pKa + log([NH3]/[NH4+]) = 9.25 + log(2.0 x 10^-4 / 0.106) = 9.70

Therefore, the pH of the resulting mixture is closest to D) 9.70.

The PH of a 250