an aqueous solution contains 0.23 m nitrous acid.
one liter of this solution could be converted into a buffer by the addition of:
(assume that the volume remains constant as each substance is added.)
(select all that apply.)
0.24 mol hcl
0.06 mol ba(oh)₂
0.11 mol hcl
0.23 mol kcl
0.24 mol kno₃

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Accepted Answer

# Brief Explanations:
To form a buffer with nitrous acid ($\ce{HNO2}$), we need a conjugate base (like $\ce{NO2^-}$) or a weak acid - strong base/strong acid - weak base reaction that leaves a mixture of weak acid and its conjugate base.
1. **For $\boldsymbol{0.06\space mol\space Ba(OH)_2}$**:
   - $\ce{Ba(OH)_2}$ is a strong base. The reaction with $\ce{HNO2}$ is $\ce{2HNO2 + Ba(OH)_2 = Ba(NO2)_2 + 2H2O}$. The moles of $\ce{OH^-}$ from $\ce{Ba(OH)_2}$ is $2	imes0.06 = 0.12\space mol$. The moles of $\ce{HNO2}$ is $0.23\space mol$. After the reaction, moles of unreacted $\ce{HNO2}=0.23 - 0.12 = 0.11\space mol$ and moles of $\ce{NO2^-}$ (from $\ce{Ba(NO2)_2}$) is $0.12\space mol$ (since 1 mole of $\ce{Ba(NO2)_2}$ gives 2 moles of $\ce{NO2^-}$, and moles of $\ce{Ba(NO2)_2}$ formed is $0.06\space mol$, so moles of $\ce{NO2^-}=2	imes0.06 = 0.12\space mol$). This gives a mixture of weak acid ($\ce{HNO2}$) and its conjugate base ($\ce{NO2^-}$), so it can form a buffer.
2. **For $\boldsymbol{0.11\space mol\space HCl}$**:
   - $\ce{HCl}$ is a strong acid. If we had a conjugate base of $\ce{HNO2}$ (like $\ce{NO2^-}$), but here we have $\ce{HNO2}$ (weak acid) and $\ce{HCl}$ (strong acid). Wait, no - actually, if we consider a reaction where we might have a conjugate base, but in this case, if we assume there was a source of $\ce{NO2^-}$ (but we don't have it initially). Wait, no, the correct way: If we have a weak acid and we add a strong acid, it won't form a buffer. Wait, I made a mistake earlier. Let's re - evaluate.
   - The correct approach: A buffer is made of a weak acid and its conjugate base or a weak base and its conjugate acid.
   - For $\ce{HNO2}$ (weak acid), to make a buffer, we can add a strong base (to partially neutralize it to form $\ce{NO2^-}$) or a salt of its conjugate base.
   - Let's re - check each option:
     - **$\boldsymbol{0.24\space mol\space HCl}$**: $\ce{HCl}$ is a strong acid. Adding it to $\ce{HNO2}$ (weak acid) will just make the solution more acidic, no conjugate base formation, so no buffer.
     - **$\boldsymbol{0.23\space mol\space KCl}$**: $\ce{KCl}$ is a neutral salt. It doesn't react with $\ce{HNO2}$ and doesn't provide a conjugate base, so no buffer.
     - **$\boldsymbol{0.24\space mol\space KNO3}$**: $\ce{KNO3}$ is a neutral salt. It doesn't react with $\ce{HNO2}$ and doesn't provide a conjugate base, so no buffer.
     - **$\boldsymbol{0.11\space mol\space HCl}$**: No, as $\ce{HCl}$ is a strong acid and doesn't form a conjugate base with $\ce{HNO2}$.
     - **$\boldsymbol{0.06\space mol\space Ba(OH)_2}$**: As corrected, the reaction between $\ce{HNO2}$ (weak acid) and $\ce{Ba(OH)_2}$ (strong base) partially neutralizes $\ce{HNO2}$ to form $\ce{NO2^-}$ (conjugate base). The moles of $\ce{OH^-}$ from $\ce{Ba(OH)_2}$ is $2	imes0.06 = 0.12\space mol$. Moles of $\ce{HNO2}=0.23\space mol$. After reaction, moles of $\ce{HNO2}$ remaining $=0.23 - 0.12 = 0.11\space mol$ and moles of $\ce{NO2^-}$ formed $= 0.12\space mol$ (from the reaction). So we have a mixture of weak acid and its conjugate base, so it can form a buffer. Also, if we consider adding a salt of the conjugate base, but in this case, the strong base reaction works.
   - Wait, another way: The buffer capacity is based on having both weak acid and conjugate base. When we add a strong base to a weak acid, we convert some of the weak acid to its conjugate base.
   - For the option $0.11\space mol\space HCl$: No, because $\ce{HCl}$ is a strong acid and will not form a conjugate base with $\ce{HNO2}$. For $0.24\space mol\space HCl$: It will just make the solution acidic, no buffer. For $0.23\space mol\space KCl$ and $0.24\space mol\space KNO3$: These are neutral salts and do not contribute to forming a conjugate base or a buffer - forming mixture with $\ce{HNO2}$.
   - The correct options are $0.06\space mol\space Ba(OH)_2$ and also, let's check $0.11\space mol\space HCl$ again. Wait, no, I think I messed up. Let's use the formula for buffer: A buffer is a solution of a weak acid ($\ce{HA}$) and its conjugate base ($\ce{A^-}$) or a weak base and its conjugate acid.
   - Nitrous acid is $\ce{HNO2}$ (weak acid). Its conjugate base is $\ce{NO2^-}$.
   - To get $\ce{NO2^-}$, we can react $\ce{HNO2}$ with a strong base (like $\ce{Ba(OH)_2}$) or a salt of $\ce{NO2^-}$.
   - When we add $0.06\space mol\space Ba(OH)_2$ (strong base), the reaction is $\ce{2HNO2 + Ba(OH)_2 = Ba(NO2)_2 + 2H2O}$. Moles of $\ce{OH^-}$ from $\ce{Ba(OH)_2}$: $n(\ce{OH^-})=2	imes0.06 = 0.12\space mol$. Moles of $\ce{HNO2}$ is $0.23\space mol$. After reaction, moles of $\ce{HNO2}$ left: $0.23 - 0.12 = 0.11\space mol$. Moles of $\ce{NO2^-}$ formed: From $\ce{Ba(NO2)_2}$, $n(\ce{NO2^-}) = 2	imes0.06=0.12\space mol$ (since 1 mole of $\ce{Ba(NO2)_2}$ gives 2 moles of $\ce{NO2^-}$). So we have a mixture of $\ce{HNO2}$ (weak acid) and $\ce{NO2^-}$ (conjugate base), so it's a buffer.
   - For $0.11\space mol\space HCl$: If we had a conjugate base (like $\ce{NO2^-}$), but we don't have it. Wait, no, the initial solution is only $\ce{HNO2}$. So adding $\ce{HCl}$ (strong acid) will not form a buffer. I was wrong earlier.
   - Another option: Let's consider the addition of a substance that can form the conjugate base. The only substance that can react with $\ce{HNO2}$ to form $\ce{NO2^-}$ (conjugate base) is a strong base. Among the options, only $\ce{Ba(OH)_2}$ is a strong base. Wait, but let's check the moles again.
   - Moles of $\ce{HNO2}=0.23\space mol$ (since the solution is $0.23\space M$ and volume is 1 L).
   - For $\ce{Ba(OH)_2}$: $n(\ce{OH^-}) = 2	imes0.06=0.12\space mol$. The reaction $\ce{HNO2 + OH^-=\ce{NO2^-}+H2O}$. So moles of $\ce{NO2^-}$ formed $ = 0.12\space mol$, moles of $\ce{HNO2}$ remaining $=0.23 - 0.12 = 0.11\space mol$. So we have both weak acid and conjugate base, so it's a buffer.
   - Also, let's check the option $0.11\space mol\space HCl$: No, because $\ce{HCl}$ is a strong acid and will not form a conjugate base with $\ce{HNO2}$. The other options ($0.24\space mol\space HCl$, $0.23\space mol\space KCl$, $0.24\space mol\space KNO3$) do not form a buffer with $\ce{HNO2}$.
   - Wait, there is another way: If we have a weak acid and we add a salt of its conjugate base. But in the given options, we don't have a salt of $\ce{NO2^-}$. So the only way is to add a strong base to partially neutralize the weak acid. So the correct option is $0.06\space mol\space Ba(OH)_2$ and also, let's re - check the $0.11\space mol\space HCl$ option. No, I think the correct answer is $0.06\space mol\space Ba(OH)_2$ and maybe I missed another. Wait, no, the initial analysis for $\ce{Ba(OH)_2}$ is correct. Also, let's consider the option $0.11\space mol\space HCl$: No, because $\ce{HCl}$ is a strong acid and $\ce{HNO2}$ is a weak acid, they can't form a buffer. The other options:
     - $0.24\space mol\space HCl$: Too much strong acid, no buffer.
     - $0.23\space mol\space KCl$: Neutral salt, no buffer.
     - $0.24\space mol\space KNO3$: Neutral salt, no buffer.
     - $0.06\space mol\space Ba(OH)_2$: Forms buffer as explained.
     - Wait, maybe I made a mistake in the first analysis of $0.11\space mol\space HCl$. Let's think again. If we had a solution of $\ce{NaNO2}$ (conjugate base) and $\ce{HNO2}$, but we don't have $\ce{NaNO2}$ here. The initial solution is only $\ce{HNO2}$. So adding $\ce{HCl}$ (strong acid) will not create a buffer. So the correct option is $0.06\space mol\space Ba(OH)_2$. Wait, but the problem says "select all that apply". Maybe I missed another. Let's re - calculate for $\ce{Ba(OH)_2}$:
       - Moles of $\ce{HNO2}=0.23\space mol$, moles of $\ce{OH^-}$ from $\ce{Ba(OH)_2}=2	imes0.06 = 0.12\space mol$. After reaction, moles of $\ce{HNO2}=0.23 - 0.12 = 0.11\space mol$, moles of $\ce{NO2^-}=0.12\space mol$ (from the reaction). So it's a buffer.
       - For $0.11\space mol\space HCl$: No, because $\ce{HCl}$ is a strong acid and $\ce{HNO2}$ is a weak acid, they can't form a buffer. So the correct answer is $0.06\space mol\space Ba(OH)_2$.
# Answer:
0.06 mol $\ce{Ba(OH)_{2}}$

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