a buffer prepared by mixing equal moles of an acid having Ka=4.5x10^-4 and a salt of its conjugate base has a pH of what?

User Icon for DrBob222 DrBob222 answered
11 years ago

See your post above.

User Icon for jo jo answered
9 years ago

420

User Icon for Explain Bot Explain Bot answered
1 year ago

To determine the pH of a buffer solution, we need to consider the equilibrium between the acid and its conjugate base. In this case, the acid and the salt of its conjugate base are mixed in equal moles. The pH of the buffer solution can be calculated using the Henderson-Hasselbalch equation:

pH = pKa + log([Salt]/[Acid])

First, let's find the pKa value. The Ka value given is 4.5x10^-4. To determine the pKa, take the negative logarithm of Ka:

pKa = -log(Ka) = -log(4.5x10^-4)

Using logarithm rules, we can simplify further:

pKa = -log(4.5) + log(10^-4) = -log(4.5) - 4

Now that we have the pKa value, we need to calculate the ratio [Salt]/[Acid]. Since the buffer is prepared using equal moles of the acid and its conjugate base, these moles cancel out, resulting in a 1:1 ratio. Therefore, [Salt]/[Acid] = 1.

Substituting the values into the Henderson-Hasselbalch equation:

pH = pKa + log(1) = pKa

Thus, the pH of the buffer solution is equal to its pKa value.