Final answer:
A buffer is formed by adding 1.0 mol of ammonia (NH3) and 1.0 mol of ammonium (NH4+) in a 1.0 L container. The pH of the diluted buffer is 9.26.
Step-by-step explanation:
A buffer solution is one that resists changes in pH when small amounts of acid or base are added to it. It is made up of a weak acid and its conjugate base, or a weak base and its conjugate acid. In the given question, a buffer is formed by adding 1.0 mol of ammonia (NH3) and 1.0 mol of ammonium (NH4+) in a 1.0 L container.
To find the pH of the diluted buffer, we can use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
where pKa is the negative logarithm of the acid dissociation constant (Ka), [A-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid. The pKa for the ammonium/ammonia buffer system is 9.26.
In this case, we have a 10.0 mL sample of the buffer which is diluted to 1000 mL with water. This means the dilution factor is 1000/10 = 100.
Therefore, the concentrations of ammonia and ammonium in the diluted buffer are [NH3] = (1.0 mol / 100) = 0.01 M and [NH4+] = (1.0 mol / 100) = 0.01 M, respectively.
Using the Henderson-Hasselbalch equation, we can plub in the values:
pH = 9.26 + log(0.01/0.01) = 9.26 + log(1) = 9.26 + 0 = 9.26
Therefore, the pH of the diluted buffer is 9.26.