Question

question 38 (1 point)
which is not a conjugate acid-base pair?
\\(\ce{h3o+}\\) and \\(\ce{h2o}\\)
\\(\ce{h2o}\\) and \\(\ce{oh-}\\)
\\(\ce{nh3}\\) and \\(\ce{nh2^{--}}\\)
\\(\ce{nh4+}\\) and \\(\ce{nh3}\\)
\\(\ce{hno2}\\) and \\(\ce{h2o}\\)

Explanation:

Step1: Recall conjugate acid - base pair definition

A conjugate acid - base pair differs by one proton ($H^+$). For example, if an acid donates a proton, it forms its conjugate base, and if a base accepts a proton, it forms its conjugate acid.

Step2: Analyze each option

- Option 1: $\ce{H3O+}$ and $\ce{H2O}$
$\ce{H3O+}$ can donate a proton to form $\ce{H2O}$ ($\ce{H3O+ \to H2O + H+}$), and $\ce{H2O}$ can accept a proton to form $\ce{H3O+}$ ($\ce{H2O + H+ \to H3O+}$). So they are a conjugate acid - base pair.
- Option 2: $\ce{H2O}$ and $\ce{OH-}$
$\ce{H2O}$ can donate a proton to form $\ce{OH-}$ ($\ce{H2O \to OH- + H+}$), and $\ce{OH-}$ can accept a proton to form $\ce{H2O}$ ($\ce{OH- + H+ \to H2O}$). So they are a conjugate acid - base pair.
- Option 3: $\ce{NH3}$ and $\ce{NH2^{2 - }}$
$\ce{NH3}$ has a nitrogen atom with a lone pair. If we consider proton transfer, $\ce{NH3}$ donating two protons would be needed to form $\ce{NH2^{2 - }}$ (since the charge changes by - 2, which is equivalent to losing two $H^+$: $\ce{NH3 \to NH2^{2 - } + 2H+}$), but a conjugate acid - base pair should differ by one proton. However, let's check other options too.
- Option 4: $\ce{NH4+}$ and $\ce{NH3}$
$\ce{NH4+}$ can donate a proton to form $\ce{NH3}$ ($\ce{NH4+ \to NH3 + H+}$), and $\ce{NH3}$ can accept a proton to form $\ce{NH4+}$ ($\ce{NH3 + H+ \to NH4+}$). So they are a conjugate acid - base pair.
- Option 5: $\ce{HNO2}$ and $\ce{H2O}$
$\ce{HNO2}$ is a weak acid, and $\ce{H2O}$ is a solvent. $\ce{HNO2}$ donates a proton to form $\ce{NO2-}$, and $\ce{H2O}$ donates a proton to form $\ce{OH-}$ or accepts a proton to form $\ce{H3O+}$. The two species $\ce{HNO2}$ and $\ce{H2O}$ do not differ by one proton. Also, comparing with option 3, but let's re - evaluate option 3. Wait, $\ce{NH3}$ to $\ce{NH2^{2 - }}$: the formula for the conjugate base of $\ce{NH3}$ should be $\ce{NH2-}$ (losing one $H^+$: $\ce{NH3 \to NH2- + H+}$), not $\ce{NH2^{2 - }}$. But the option with $\ce{HNO2}$ and $\ce{H2O}$: $\ce{HNO2}$ and $\ce{H2O}$ have no proton - transfer relationship to be a conjugate pair. Wait, maybe I made a mistake earlier. Let's re - check:

Wait, the key is that a conjugate acid - base pair must differ by exactly one $H^+$. Let's check the charge and proton difference:

- For $\ce{H3O+}$ ($\ce{H2O + H+}$) and $\ce{H2O}$: differ by 1 $H^+$.
- $\ce{H2O}$ ($\ce{OH- + H+}$) and $\ce{OH-}$: differ by 1 $H^+$.
- $\ce{NH4+}$ ($\ce{NH3 + H+}$) and $\ce{NH3}$: differ by 1 $H^+$.
- $\ce{NH3}$ and $\ce{NH2^{2 - }}$: $\ce{NH3}$ has a charge of 0, $\ce{NH2^{2 - }}$ has a charge of - 2. The difference in $H^+$: to go from $\ce{NH3}$ (0 charge) to $\ce{NH2^{2 - }}$ (- 2 charge), we need to lose 2 $H^+$ (since each $H^+$ is + 1, losing 2 $H^+$ would change charge by - 2: $0-2\times( + 1)=-2$). But a conjugate pair should lose 1 $H^+$. However, the option $\ce{HNO2}$ and $\ce{H2O}$: $\ce{HNO2}$ has formula $\ce{HNO2}$, $\ce{H2O}$ is $\ce{H2O}$. The two molecules have different atoms (N vs O in the main part). The conjugate acid of $\ce{H2O}$ is $\ce{H3O+}$, conjugate base is $\ce{OH-}$. The conjugate base of $\ce{HNO2}$ is $\ce{NO2-}$. So $\ce{HNO2}$ and $\ce{H2O}$ are not a conjugate acid - base pair. Also, the option $\ce{NH3}$ and $\ce{NH2^{2 - }}$: maybe a typo, but the most clear non - conjugate pair is $\ce{HNO2}$ and $\ce{H2O}$. Wait, no, let's check the options again. The options are:

1. $\ce{H3O+}$ and $\ce{H2O}$: conjugate (acid - base: $\ce{H3O+}$ is acid, $\ce{H2O}$ is base)
2. $\ce{H2O}$ and $\ce{OH-}$: conjugate (acid - base: $\ce{H2O}$ is acid, $\ce{OH-}$ is base)
3. $\ce{NH3}$ and $\ce{NH2^{2 - }}$: not conjugate (should differ by 1 $H^+$, here differs by 2 $H^+$? Wait, $\ce{NH3}$ has 3 H, $\ce{NH2^{2 - }}$ has 2 H, so difference of 1 H, but charge: $\ce{NH3}$ is 0, $\ce{NH2^{2 - }}$ is - 2. The charge difference: when you lose 1 $H^+$, charge should decrease by 1 (since $H^+$ is + 1). So $\ce{NH3}$ (0) losing 1 $H^+$ should be $\ce{NH2-}$ (- 1), not $\ce{NH2^{2 - }}$ (- 2). So this is an error in the formula, but the option $\ce{HNO2}$ and $\ce{H2O}$: $\ce{HNO2}$ and $\ce{H2O}$ have no proton - transfer relation to be conjugate. So the answer is $\ce{HNO2}$ and $\ce{H2O}$, i.e., the option with $\ce{HNO2}$ and $\ce{H2O}$.

Answer:

The option with $\ce{HNO2}$ and $\ce{H2O}$ (the last option: $\boldsymbol{\ce{HNO2}}$ and $\boldsymbol{\ce{H2O}}$)