Question

use the following information to answer the next question.
the following are examples of acid - base reactions:
(i) $\ce{nh_{3}(aq) + h_{2}o(l) \
ightarrow nh_{4}^{+}(aq) + oh^{-}(aq)}$
(ii) $\ce{h_{2}o(l) + ch_{3}cooh(aq) \
ightarrow h_{3}o^{+}(aq) + ch_{3}coo^{-}(aq)}$
(iii) $\ce{h_{2}o(l) + co_{3}^{2 - }(aq) \
ightarrow hco_{3}^{-}(aq) + oh^{-}(aq)}$
in which of the above reactions does water act as a brønsted - lowry acid:
\bigcirc (i) and (ii)
\bigcirc (ii) and (iii)
\bigcirc (iii) only
\bigcirc (i) and (iii)
\bigcirc (ii) only
question 26 (1 point)
use the following information to answer the next question.
bicarbonate is an ion with the chemical formula $\ce{hco_{3}^{-}}$. in the presence of water, it results in the
following reactions:
(i) $\ce{hco_{3}^{-}(aq) + h_{2}o(l) \
ightarrow h_{2}co_{3}(aq) + oh^{-}(aq)}$
(ii) $\ce{hco_{3}^{-}(aq) + h_{2}o(l) \
ightarrow co_{3}^{2 - }(aq) + h_{3}o^{+}(aq)}$
in the above reactions, the bicarbonate ion acts as ______.
\bigcirc a strong acid
\bigcirc a strong base
\bigcirc an acid in reaction (i) and as a base in reaction (ii)
\bigcirc a base in reaction (i) and as an acid in reaction (ii)
\bigcirc a weak base

Explanation:

Question 25

Step1: Recall Brønsted - Lowry acid definition

A Brønsted - Lowry acid is a proton ($H^+$) donor.

Step2: Analyze reaction (i)

In reaction (i): $\ce{NH3(aq) + H2O(l) -> NH4^{+}(aq) + OH^{-}(aq)}$, water donates a proton? No, water accepts a proton? Wait, $\ce{NH3}$ is a base (accepts $H^+$), water donates $H^+$? Wait, $\ce{NH3 + H2O -> NH4+ + OH-}$, here $\ce{H2O}$ donates $H^+$ to $\ce{NH3}$? Wait, no, $\ce{NH3}$ is a base (Lewis base, but Brønsted - Lowry base accepts $H^+$). Wait, $\ce{H2O}$ in reaction (i): $\ce{H2O}$ gives $H^+$ to $\ce{NH3}$ to form $\ce{NH4+}$, so $\ce{H2O}$ is an acid? Wait, no, wait the products: $\ce{OH-}$ is formed. Wait, maybe I made a mistake. Let's re - analyze.

Brønsted - Lowry acid: donates $H^+$, Brønsted - Lowry base: accepts $H^+$.

Reaction (i): $\ce{NH3(aq) + H2O(l) -> NH4^{+}(aq) + OH^{-}(aq)}$

$\ce{NH3}$ accepts $H^+$ (from $\ce{H2O}$) to form $\ce{NH4+}$, so $\ce{NH3}$ is a base. $\ce{H2O}$ donates $H^+$? But it forms $\ce{OH-}$, which is a base. Wait, no, in this reaction, $\ce{H2O}$ is acting as an acid? Wait, no, let's look at the proton transfer. $\ce{H2O}$ has $H^+$ to donate, $\ce{NH3}$ accepts it. So $\ce{H2O}$ is an acid here? Wait, but in reaction (ii):

Reaction (ii): $\ce{H2O(l) + CH3COOH(aq) -> H3O^{+}(aq) + CH3COO^{-}(aq)}$

$\ce{CH3COOH}$ donates $H^+$ to $\ce{H2O}$, so $\ce{CH3COOH}$ is an acid, $\ce{H2O}$ accepts $H^+$ to form $\ce{H3O+}$, so $\ce{H2O}$ is a base here.

Reaction (iii): $\ce{H2O(l) + CO3^{2 - }(aq) -> HCO3^{-}(aq) + OH^{-}(aq)}$

$\ce{CO3^{2 - }}$ accepts $H^+$ from $\ce{H2O}$ to form $\ce{HCO3-}$, so $\ce{CO3^{2 - }}$ is a base, $\ce{H2O}$ donates $H^+$ (since it forms $\ce{OH-}$ after losing $H^+$), so $\ce{H2O}$ is an acid here.

Wait, now I'm confused. Let's re - define:

Brønsted - Lowry acid: species that donates a proton ($H^+$).

Brønsted - Lowry base: species that accepts a proton ($H^+$).

Reaction (i):

$\ce{NH3 + H2O -> NH4+ + OH-}$

$\ce{NH3}$ (base) accepts $H^+$ from $\ce{H2O}$ (acid) to form $\ce{NH4+}$. $\ce{H2O}$ (acid) donates $H^+$ to form $\ce{OH-}$.

Reaction (ii):

$\ce{H2O + CH3COOH -> H3O+ + CH3COO-}$

$\ce{CH3COOH}$ (acid) donates $H^+$ to $\ce{H2O}$ (base) to form $\ce{H3O+}$.

Reaction (iii):

$\ce{H2O + CO3^{2 - } -> HCO3- + OH-}$

$\ce{CO3^{2 - }}$ (base) accepts $H^+$ from $\ce{H2O}$ (acid) to form $\ce{HCO3-}$. $\ce{H2O}$ (acid) donates $H^+$ to form $\ce{OH-}$.

Wait, so in reaction (i) and (iii), water donates $H^+$ (acts as Brønsted - Lowry acid), and in reaction (ii) water accepts $H^+$ (acts as Brønsted - Lowry base). Wait, but the options are:

(i) and (ii)

(ii) and (iii)

(iii) only

(i) and (iii)

(ii) only

Wait, maybe my initial analysis was wrong. Let's check again.

Reaction (i): $\ce{NH3 + H2O -> NH4+ + OH-}$

$\ce{NH3}$ is a base (accepts $H^+$), $\ce{H2O}$ is an acid (donates $H^+$) because it gives $H^+$ to $\ce{NH3}$.

Reaction (ii): $\ce{H2O + CH3COOH -> H3O+ + CH3COO-}$

$\ce{CH3COOH}$ is an acid (donates $H^+$), $\ce{H2O}$ is a base (accepts $H^+$) because it takes $H^+$ from $\ce{CH3COOH}$.

Reaction (iii): $\ce{H2O + CO3^{2 - } -> HCO3- + OH-}$

$\ce{CO3^{2 - }}$ is a base (accepts $H^+$), $\ce{H2O}$ is an acid (donates $H^+$) because it gives $H^+$ to $\ce{CO3^{2 - }}$.

So water acts as Brønsted - Lowry acid in (i) and (iii). So the correct option is (i) and (iii).

Step1: Recall Brønsted - Lowry acid and base definitions

A Brønsted - Lowry acid donates a proton ($H^+$), a Brønsted - Lowry base accepts a proton ($H^+$).

Step2: Analyze reaction (i)

Reaction (i): $\ce{HCO3^{-}(aq) + H2O(l) -> H2CO3(aq) + OH^{-}(aq)}$

$\ce{HCO3-}$ accepts a proton from $\ce{H2O}$ (since $\ce{H2O}$ donates $H^+$ to form $\ce{OH-}$? Wait, no: $\ce{HCO3- + H2O -> H2CO3 + OH-}$. $\ce{HCO3-}$ gains a proton (from $\ce{H2O}$) to form $\ce{H2CO3}$, so $\ce{HCO3-}$ is a base (accepts $H^+$).

Step3: Analyze reaction (ii)

Reaction (ii): $\ce{HCO3^{-}(aq) + H2O(l) -> CO3^{2 - }(aq) + H3O^{+}(aq)}$

$\ce{HCO3-}$ donates a proton to $\ce{H2O}$ (since $\ce{H2O}$ accepts $H^+$ to form $\ce{H3O+}$), so $\ce{HCO3-}$ is an acid (donates $H^+$).

So bicarbonate ion acts as a base in reaction (i) and as an acid in reaction (ii).

Answer:

(i) and (iii)

Question 26