Q19.4P - Chemistry: Molecular Nature Of Matter And Change

Question

The scenes below depict solutions of the same HA/A buffer (with counterions and water molecules omitted for clarity). 

(a) Which solution has the greatest buffer capacity?

 (b) Explain how the pH ranges of the buffers compare. 

(c) Which solution can react with the largest amount of added strong acid?

Step-by-Step Solution

Answer

(a) Beaker 3 has the greatest buffer capacity

(b) 4 beakers are in the effective pH range for buffers

(c) Beaker 2 with the greatest amount of ${\text{A}}^{\text{ - }}$ in the solution will react with the acid.

Step 1: To find the solution has the greatest buffer capacity

(a)

The buffer capacity of a solution is when its ${\text{pH = pK}}_{\text{a}}$. We know that ${\text{pH = pK}}_{\text{a}}\text{ + log}\left(\frac{\left[{\text{A}}^{\text{ - }}\right]}{\text{[HA]}}\right){\text{.So for pH = pK}}_{\text{a}}\text{,log}\left(\frac{\left[{\text{A}}^{\text{ - }}\right]}{\text{[[HA]}}\right)$ must be equal to 0 . $\text{log(1) = 0.Therefore,}\left(\frac{\left[{\text{A}}^{\text{ - }}\right]}{\text{[HA]}}\right)\text{ = 1}$.

So for a solution to have a solution to have a greatest buffer capacity, $\frac{\left[{\text{A}}^{\text{ - }}\right]}{\text{[HA]}}$ must be the nearest to 1 . Evaluate each beaker.

Beaker 1

$\text{[HA] = 7/}\left[{\text{A}}^{\text{ - }}\right]\text{ = 3}\frac{\left[{\text{A}}^{\text{ - }}\right]}{\text{[HA]}}\text{ = }\frac{\text{[3]}}{\text{[7]}}\text{ = 0.429}$

Beaker 2

$\text{[HA] = 3/}\left[{\text{A}}^{\text{ - }}\right]\text{ = 7}\frac{\left[{\text{A}}^{\text{ - }}\right]}{\text{[HA]}}\text{ = }\frac{\text{[7]}}{\text{[3]}}\text{ = 2.333}$

Beaker 3

$\text{[HA] = 5}\left[{\text{A}}^{\text{ - }}\right]\text{ = 5}\frac{\left[{\text{A}}^{\text{ - }}\right]}{\text{[HA]}}\text{ = }\frac{\text{[5]}}{\text{[5]}}\text{ = 1}$

Beaker 4

$\text{[HA] = 3/}\left[{\text{A}}^{\text{ - }}\right]\text{ = 3}\frac{\left[{\text{A}}^{\text{ - }}\right]}{\text{[HA]}}\text{ = }\frac{\text{[3]}}{\text{[3]}}\text{ = 1}$

It can be observed that beaker 3 and 4 both have ratio between [HA] and $\left[{\text{A}}^{\text{ - }}\right]$ Another thing that can affect the buffer capacity is the concentration. A more concentrated solution will have a greater buffer capacity. Beaker 3 is more concentrated than Beaker 4 .

Therefore, Beaker 3 has the greatest buffer capacity.

Step 2:To find how the pH ranges of the buffers compare

(b)

Solve the pH of each beaker using the Henderson-Hasslebalch equation.

Beaker 1:

$$\begin{gathered} {\text{pH = pK}}_{\text{a}}\text{ + log}\frac{\left[{\text{A}}^{\text{ - }}\right]}{\text{[HA]}}{\text{pK}}_{\text{a}}\text{ + log}\frac{\text{3}}{\text{7}} \\ {\text{pH = pK}}_{\text{a}}\text{ - 0.368} \end{gathered}$$

Beaker 2:

$$\begin{gathered} {\text{pH = pK}}_{\text{a}}\text{ + log}\frac{\left[{\text{A}}^{\text{ - }}\right]}{\text{[HA]}}{\text{pK}}_{\text{a}}\text{ + log}\frac{\text{7}}{\text{2}} \\ {\text{pH = pK}}_{\text{a}}\text{ + 0.368} \end{gathered}$$

Beaker 3:

$$\begin{gathered} {\text{pH = pK}}_{\text{a}}\text{ + log}\frac{\left[{\text{A}}^{\text{ - }}\right]}{\text{[HA]}}{\text{pK}}_{\text{a}}\text{ + log}\frac{\text{5}}{\text{5}} \\ {\text{pH = pK}}_{\text{a}} \end{gathered}$$

Beaker 4:

$$\begin{gathered} {\text{pH = pK}}_{\text{a}}\text{ + log}\frac{\left[{\text{A}}^{\text{ - }}\right]}{\text{[HA]}}{\text{pK}}_{\text{a}}\text{ + log}\frac{\text{3}}{\text{3}} \\ {\text{pH = pK}}_{\text{a}} \end{gathered}$$

All 4 beakers are in the effective pH range for buffers

Step 3: To find the solution can react with the largest amount of added strong acid

(c)

The solution that can react with the largest amount ${\text{A}}^{\text{ - }}$ can react with the largest amount of strong acid. Strong acid will dissociate completely. The presence of ${\text{A}}^{\text{ - }}$ will react with the ${\text{H}}_3{\text{O}}^{+}$ produced by the strong acid.

${\text{H}}_3{\text{O}}^{+}+{\text{A}}^{-}\rightleftharpoons {\text{H}}_2{\text{O}}^{+}+\text{HA}$

Therefore, Beaker 2 with the greatest amount of ${\text{A}}^{\text{ - }}$ in the solution will react with the acid.