Question

A 150.0 ml buffer solution is 0.110 moll−1 in ch3nh2 and 0.135 moll−1 in ch3nh3br .if the same volume of the buffer were 0.250 moll−1 in ch3nh2 and 0.400 moll−1 in ch3nh3br , what mass of hcl could be handled before the ph fell below 9.90?

Answer 1

The mass of HCl that can be handled before the pH falls below 9.90 cannot be determined because the concentrations of [A-] and [HA] would result in a negative value for [HA].

Step-by-step explanation:

To solve this problem, we can use the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the concentration of its components. The Henderson-Hasselbalch equation is:

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

In this case, the weak base is CH3NH2 and its conjugate acid is CH3NH3+. The pKa of CH3NH2 is given as 9.25. We can rearrange the Henderson-Hasselbalch equation as:

[A-]/[HA] = 10^(pH - pKa)

Now let's use the given information to solve the problem:

Given:

• Initial volume of buffer solution = 150.0 ml
• [CH3NH2] = 0.110 M
• [CH3NH3+] = 0.135 M
• New volume of buffer solution = 150.0 ml
• [CH3NH2] = 0.250 M
• [CH3NH3+] = 0.400 M
• pH threshold = 9.90

First, we need to calculate the initial ratio of [A-]/[HA] using the Henderson-Hasselbalch equation:

[A-]/[HA] = 10^(pH - pKa) = 10^(9.90 - 9.25) = 4.765

Since the new volume of the buffer solution is the same as the initial volume, the volume ratio of [A-]/[HA] remains the same. Therefore, the ratio of [A-]/[HA] is still 4.765.

We can use this ratio to calculate the new concentrations of [A-] and [HA]:

[A-] = [A-]/[HA] * [HA] = 4.765 * 0.110 M = 0.519 M

[HA] = 0.135 M - [A-] = 0.135 M - 0.519 M = -0.384 M

As [HA] cannot be negative, we can conclude that the pH will fall below the threshold before any HCl can be handled by the buffer solution.