Question

question 10 (1 point)
which of the following statements is correct?
○ ch₃coo⁻(aq) is a stronger conjugate base of weak acid ch₃cooh(aq)
○ ch₃coo⁻(aq) is a weaker conjugate base of weak acid ch₃cooh (aq)
○ ch₃coo⁻(aq) is a stronger conjugate base of strong acid ch₃cooh(aq)
○ ch₃coo⁻(aq) is a weaker conjugate base of strong acid ch₃cooh(aq)
○ ch₃coo⁻(aq) is a stronger conjugate base of weak base ch₃cooh(aq)

question 11 (1 point)
what is the concentration of oh⁻ if an hcl solution has a concentration of 0.07 mol/l?
○ 1.15
○ 7 × 10⁻¹²
○ 1.285 × 10¹
○ 1.43 × 10⁻¹³

question 12 (1 point)
if an acid has a kₐ of 4.1 × 10⁻⁹ then what can we assume is the source of h₃o⁺ ions?
○ the source of ions is from the acid.
○ the source of ions is from the water.
○ the ions come from both water and acids.
○ the ions come from both conjugate acid and base pairs.

Explanation:

Question 10

To determine the correct statement, we analyze the conjugate acid - base pair. Acetic acid ($\ce{CH_{3}COOH}$) is a weak acid (it does not dissociate completely in water). For a conjugate acid - base pair, the weaker the acid, the stronger its conjugate base. The conjugate base of $\ce{CH_{3}COOH}$ is $\ce{CH_{3}COO^{-}}$.

• Option 1: $\ce{CH_{3}COO^{-}(aq)}$ is a stronger conjugate base of weak acid $\ce{CH_{3}COOH(aq)}$ - This is correct because weak acids have strong conjugate bases.
• Option 2: Claims $\ce{CH_{3}COO^{-}}$ is a weaker conjugate base, which is wrong as weak acids have strong conjugate bases.
• Option 3: $\ce{CH_{3}COOH}$ is not a strong acid, so this is wrong.
• Option 4: $\ce{CH_{3}COOH}$ is not a strong acid, so this is wrong.
• Option 5: $\ce{CH_{3}COOH}$ is an acid, not a base, so this is wrong.

Step 1: Recall the ion - product of water

The ion - product of water, $K_{w}=[\ce{H^{+}}][\ce{OH^{-}}] = 1.0\times10^{- 14}$ at $25^{\circ}\text{C}$. HCl is a strong acid, so it dissociates completely in water: $\ce{HCl(aq)
ightarrow H^{+}(aq) + Cl^{-}(aq)}$. So, $[\ce{H^{+}}]=[\ce{HCl}]=0.07\space mol/L$.

Step 2: Calculate $[\ce{OH^{-}}]$

We can rearrange the $K_{w}$ expression to solve for $[\ce{OH^{-}}]$: $[\ce{OH^{-}}]=\frac{K_{w}}{[\ce{H^{+}}]}$.
Substitute $K_{w} = 1.0\times10^{-14}$ and $[\ce{H^{+}}]=0.07\space mol/L$ into the formula:
$[\ce{OH^{-}}]=\frac{1.0\times 10^{-14}}{0.07}\approx1.43\times 10^{-13}\space mol/L$

The acid - dissociation constant $K_{a}$ of the acid is $4.1\times 10^{-9}$, which is a very small value. This means that the acid is a very weak acid and dissociates to a very small extent, producing a negligible amount of $\ce{H_{3}O^{+}}$ ions. In aqueous solutions, even in the presence of a weak acid, the main source of $\ce{H_{3}O^{+}}$ ions when the acid is very weak (with a very small $K_{a}$) is the auto - ionization of water.

• Option 1: The acid is very weak, so it does not produce a significant amount of $\ce{H_{3}O^{+}}$ ions, so this is wrong.
• Option 2: The source of ions is from the water - This is correct because the acid is so weak that its contribution to $\ce{H_{3}O^{+}}$ is negligible compared to the auto - ionization of water.
• Option 3: The acid does not contribute significantly, so ions do not come from both water and acid.
• Option 4: The question is about the source of $\ce{H_{3}O^{+}}$ ions, not conjugate acid - base pairs, so this is wrong.

Answer:

A. $\ce{CH_{3}COO^{-}(aq)}$ is a stronger conjugate base of weak acid $\ce{CH_{3}COOH(aq)}$

Question 11