Q18.154CP - Chemistry: Molecular Nature Of Matter And Change

Question

The beakers shown contain 0.300 Lof aqueous solutions of a moderately weak acid HY. Each particle represents 0.010 mol; solvent molecules are omitted for clarity. 

(a) The reaction in beaker Ais at equilibrium. Calculate Q for B, C, and D to determine which, if any, is also at equilibrium. 

(b) For any not at equilibrium, in which direction does the reaction proceed? 

(c) Does dilution affect the extent of dissociation of a weak acid? Explain.

Step-by-Step Solution

Answer

(a) The K of beaker A and the Q of beaker D are approximately equal. Therefore, the solution in beaker D is also in equilibrium.

(b)In backward direction ,the reaction will proceed.

(c) Dilution increases the extent of dissociation of a weak acid.

Step 1: To determine which, if any, is also at equilibrium.

(a)

The green spheres represent Y, white spheres represent, and red spheres represent O.First, solve for the concentration of each chemical species.

There are ${\text{4Y}}^{\text{-}}$atoms. So

$\begin{array}{l}\left[{\text{Y}}^{\text{-}}\right]\text{=}\frac{\text{4×0.010 mol}}{\text{0.300 L}}\\ \left[{\text{Y}}^{\text{-}}\right]\text{=0.133M}\end{array}$

There are data-custom-editor="chemistry" $4{\text{H}}_3{\text{O}}^{+}$atoms. So...

data-custom-editor="chemistry" $\begin{array}{l}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{ =}\frac{\text{4×0.010 mol}}{\text{0.300 L}}\\ \left[{\text{Y}}^{\text{-}}\right]\text{ =0.133}\end{array}$

There are atoms. So...

data-custom-editor="chemistry" $\begin{array}{l}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{ \&=}\frac{\text{8×0.010 mol}}{\text{0.300 L}}\\ \left[{\text{Y}}^{\text{-}}\right]\text{ =0.267M}\end{array}$

Now, solve for the K of the dissociation of HY in water.

$\begin{array}{l}\text{K=}\frac{\left[{\text{Y}}^{\text{-}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}{\text{[HY]}}\\ \text{=}\frac{\text{(0.133)(0.133)}}{\text{0.267}}\\ {\text{K =6.63×10}}^{\text{-2}}\end{array}$

Step 2: To solve for the Q of the Beaker B

Next, solve for the Q of the other beakers. Beaker B

There are $\text{2Y atoms can}$

data-custom-editor="chemistry" $\begin{array}{l}\left[{\text{Y}}^{\text{-}}\right]\text{=}\frac{\text{2×0.010 mol}}{\text{0.300 L}}\\ \left[{\text{Y}}^{\text{-}}\right]\text{=0.0667M}\end{array}$

There are data-custom-editor="chemistry" ${\text{2H}}_{\text{3}}{\text{O}}^{\text{+}}\text{atoms}$So

data-custom-editor="chemistry" $\begin{array}{l}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{ =}\frac{\text{2×0.010 mol}}{\text{0.300 L}}\\ \left[{\text{Y}}^{\text{-}}\right]\text{ =0.0667M}\end{array}$

There are 6HY atoms. So..

data-custom-editor="chemistry" $\begin{array}{l}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{ =}\frac{\text{6×0.010 mol}}{\text{0.300 L}}\\ \left[{\text{Y}}^{\text{-}}\right]\text{ =0.200M}\end{array}$

Now, solve for the Q of the dissociation of HY in water.

data-custom-editor="chemistry" $\begin{array}{l}{\text{Q}}_{\text{B}}\text{ =}\frac{\left[{\text{Y}}^{\text{-}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}{\text{[HY]}}\\ \text{=}\frac{\text{(0.0667)(0.0667)}}{\text{0.200}}\\ {\text{Q}}_{\text{B}}{\text{ =2.22×10}}^{\text{-2}}\end{array}$

Step 3: To solve for the Q of the Beaker C

Beaker C

There are ${\text{2Y}}^{\text{-}}$atoms. So...

$\begin{array}{l}\left[{\text{Y}}^{\text{-}}\right]\text{=}\frac{\text{2×0.010 mol}}{\text{0.300 L}}\\ \left[{\text{Y}}^{\text{-}}\right]\text{=0.0667M}\end{array}$

There are$2{\text{H}}_3{\text{O}}^{+}$ atoms. SO...

$\begin{array}{l}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{ =}\frac{\text{2×0.010 mol}}{\text{0.300 L}}\\ \left[{\text{Y}}^{\text{-}}\right]\text{ =0.0667M}\end{array}$

There are $\text{4HY}$atoms. So...

$\begin{array}{l}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{ =}\frac{\text{4×0.010 mol}}{\text{0.300 L}}\\ \left[{\text{Y}}^{\text{-}}\right]\text{ =0.133M}\end{array}$

Now, solve for the Q of the dissociation of HY in water.

$\begin{array}{l}{\text{Q}}_{\text{C}}\text{ =}\frac{\left[{\text{Y}}^{\text{-}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}{\text{[HY]}}\\ \text{=}\frac{\text{(0.0667)(0.0667)}}{\text{0.133}}\\ {\text{Q}}_{\text{C}}{\text{ =3.35×10}}^{\text{-2}}\end{array}$

Step 4: To solve for the Q of the Beaker D

Beaker D

There are${\text{2Y}}^{\text{-}}$ atoms. So...

$\begin{array}{l}\left[{\text{Y}}^{\text{-}}\right]\text{=}\frac{\text{2×0.010 mol}}{\text{0.300 L}}\\ \left[{\text{Y}}^{\text{-}}\right]\text{=0.0667M}\end{array}$

There are$2{\text{H}}_3{\text{O}}^{+}$ atoms. So...

$\begin{array}{l}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{ =}\frac{\text{2×0.010 mol}}{\text{0.300 L}}\\ \left[{\text{Y}}^{\text{-1}}\right]\text{ =0.0667}\end{array}$

There are $\text{2HY}$atoms. So...

$\begin{array}{l}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{ =}\frac{\text{2×0.010 mol}}{\text{0.300 L}}\\ \left[{\text{Y}}^{\text{-1}}\right]\text{ =0.0667}\end{array}$

Now, solve for the Q of the dissociation of HY in water.

$\begin{array}{l}{\text{Q}}_{\text{D}}\text{ =}\frac{\left[{\text{Y}}^{\text{-}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}{\text{[HY]}}\\ \text{=}\frac{\text{(0.0667)(0.0667)}}{\text{0.0667}}\end{array}$

${\text{Q}}_{\text{D}}{\text{=6.67×10}}^{\text{-2}}$

Hence the K of beaker A and the Q of beaker D are approximately equal. Therefore, the solution in beaker D is also in equilibrium.

Step 5:To find the direction

(b)

Recall the Q of each beaker and the equilibrium constant K :

$\begin{array}{l}{\text{K=6.63×10}}^{\text{-2}}\\ {\text{Q}}_{\text{B}}{\text{=2.22×10}}^{\text{-2}}\\ {\text{Q}}_{\text{C}}{\text{=3.35×10}}^{\text{-2}}\\ {\text{Q}}_{\text{D}}{\text{=6.67×10}}^{\text{-2}}\end{array}$

Now let us evaluate each.

Beaker B: Since,${\text{K>Q}}_{\text{B}}$ the reaction will proceed forward.

Beaker C: Since,${\text{K>Q}}_{\text{B}}$ the reaction will proceed forward

Step 6: To find the dilution affect the extent of dissociation of a weak acid

(c)

Based on the photos provided, the Q of reaction increases as lesser particles exist. Lesser particles in a beaker with equal volumes also means that it is more dilute. The Q of the reaction approaches the value of the equilibrium constant K.

 

Therefore, dilution affect the dissociation of weak acids. It increases the extent of dissociation of a weak acid.