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A buffer is formed by adding 1.0 mol of ammonia, nh3, and 1.0 mol ammonium, nh4 , in a 1.0 l container. If a 10.0 ml sample of this buffer is diluted to 1000 ml with water, what is the pH of the diluted buffer?

User Ftexperts
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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.

User Red Swan
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