Question

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 kno₂

Explanation:

To form a buffer with nitrous acid ($\ce{HNO_2}$), we need a conjugate base (like $\ce{NO_2^-}$) or a strong base that partially neutralizes $\ce{HNO_2}$ to form the conjugate base (without completely consuming the acid). Let's analyze each option:

Step 1: Analyze 0.24 mol HCl

HCl is a strong acid. Adding it to $\ce{HNO_2}$ (a weak acid) will increase $\ce{H^+}$ but not form a conjugate base - it will suppress dissociation, not form a buffer. So this is incorrect.

Step 2: Analyze 0.06 mol $\ce{Ba(OH)_2}$

$\ce{Ba(OH)_2}$ is a strong base. The reaction is: $\ce{2HNO_2 + Ba(OH)_2 -> Ba(NO_2)_2 + 2H_2O}$. Moles of $\ce{OH^-}$ from $\ce{Ba(OH)_2}$: $0.06\ \text{mol} \times 2 = 0.12\ \text{mol}$. Moles of $\ce{HNO_2}$ initially: $0.23\ \text{mol}$ (from $0.23\ \text{M}$ in $1\ \text{L}$). After reaction, moles of $\ce{HNO_2}$ remaining: $0.23 - 0.12 = 0.11\ \text{mol}$, and moles of $\ce{NO_2^-}$ formed: $0.12\ \text{mol}$ (from the reaction, since 2 moles $\ce{HNO_2}$ give 1 mole $\ce{Ba(NO_2)_2}$, so moles of $\ce{NO_2^-}$ is $0.12\ \text{mol}$). Now we have a weak acid ($\ce{HNO_2}$) and its conjugate base ($\ce{NO_2^-}$) - a buffer.

Step 3: Analyze 0.11 mol HCl

HCl is a strong acid. It will react with any conjugate base, but we don't have a conjugate base initially. Adding HCl to $\ce{HNO_2}$ will just make the solution more acidic, no buffer. Incorrect.

Step 4: Analyze 0.23 mol KCl

KCl is a neutral salt (from strong acid and strong base). It doesn't react with $\ce{HNO_2}$ or provide a conjugate base. So it won't form a buffer. Incorrect.

Step 5: Analyze 0.24 mol $\ce{KNO_2}$

$\ce{KNO_2}$ dissociates to give $\ce{NO_2^-}$ (conjugate base of $\ce{HNO_2}$). Now we have $\ce{HNO_2}$ (weak acid) and $\ce{NO_2^-}$ (conjugate base) - this is a buffer (weak acid + conjugate base).

Wait, wait, initial moles of $\ce{HNO_2}$: $0.23\ \text{mol}$ (1 L of 0.23 M). Let's re - check the $\ce{KNO_2}$ option:

Moles of $\ce{NO_2^-}$ from $\ce{KNO_2}$: $0.24\ \text{mol}$. Moles of $\ce{HNO_2}$: $0.23\ \text{mol}$. So we have a weak acid and its conjugate base (even though the conjugate base is in slight excess), this is a buffer.

Wait, earlier analysis of $\ce{Ba(OH)_2}$:

Moles of $\ce{HNO_2}$: $0.23\ \text{mol}$ (1 L * 0.23 M). Moles of $\ce{OH^-}$ from $\ce{Ba(OH)_2}$: $0.06\ \text{mol} \times 2 = 0.12\ \text{mol}$. The reaction is $\ce{HNO_2 + OH^- -> NO_2^- + H_2O}$. So moles of $\ce{HNO_2}$ remaining: $0.23 - 0.12 = 0.11\ \text{mol}$, moles of $\ce{NO_2^-}$ formed: $0.12\ \text{mol}$. So we have $\ce{HNO_2}$ (0.11 mol) and $\ce{NO_2^-}$ (0.12 mol) - buffer.

For $\ce{KNO_2}$:

Moles of $\ce{NO_2^-}$: $0.24\ \text{mol}$, moles of $\ce{HNO_2}$: $0.23\ \text{mol}$. So weak acid and conjugate base - buffer.

Wait, but let's check the initial problem again. The initial solution is 0.23 M $\ce{HNO_2}$ in 1 L, so 0.23 mol $\ce{HNO_2}$.

Wait, maybe I made a mistake with $\ce{Ba(OH)_2}$:

The reaction is $\ce{2HNO_2 + Ba(OH)_2 -> Ba(NO_2)_2 + 2H_2O}$. So 1 mole $\ce{Ba(OH)_2}$ reacts with 2 moles $\ce{HNO_2}$. Moles of $\ce{Ba(OH)_2}$: 0.06 mol, so moles of $\ce{HNO_2}$ reacted: $0.06 \times 2 = 0.12$ mol. Moles of $\ce{HNO_2}$ remaining: $0.23 - 0.12 = 0.11$ mol. Moles of $\ce{NO_2^-}$ formed: $0.12$ mol (since 2 moles $\ce{HNO_2}$ produce 2 moles $\ce{NO_2^-}$? Wait, no: $\ce{Ba(NO_2)_2}$ dissociates into $\ce{Ba^{2+}}$ and $2\ce{NO_2^-}$. So moles of $\ce{NO_2^-}$ from $\ce{Ba(NO_2)_2}$: $0.06\ \text{mol} \times 2 = 0.12\ \text{mol}$. So now we have $\ce{HNO_2}$ (0.11 mol) and $\ce{NO_2^-}$ (0.12 mol) - buffer.

For $\ce{KNO_2}$: $\ce{KNO_2}$ dissociates into $\ce{K^+}$ and $\ce{NO_2^-}$. So moles of $\ce{NO_2^-}$: 0.24 mol, moles of $\ce{HNO_2}$: 0.23 mol. So weak acid and conjugate base - buffer.

Wait, but the option 0.24 mol HCl is wrong, 0.11 mol HCl is wrong, 0.23 mol KCl is wrong. Now between 0.06 mol $\ce{Ba(OH)_2}$ and 0.24 mol $\ce{KNO_2}$.

Wait, let's check the buffer definition: a buffer is a solution containing a weak acid and its conjugate base (or weak base and its conjugate acid) in significant amounts.

For $\ce{Ba(OH)_2}$:

- Weak acid: $\ce{HNO_2}$ (0.11 mol)
- Conjugate base: $\ce{NO_2^-}$ (0.12 mol) → buffer.

For $\ce{KNO_2}$:

- Weak acid: $\ce{HNO_2}$ (0.23 mol)
- Conjugate base: $\ce{NO_2^-}$ (0.24 mol) → buffer.

Wait, but let's check the moles again. The initial concentration is 0.23 M in 1 L, so moles of $\ce{HNO_2}$ is 0.23 mol.

Wait, maybe I messed up the $\ce{Ba(OH)_2}$ reaction. Let's do it again:

$\ce{HNO_2 + OH^- -> NO_2^- + H_2O}$

Moles of $\ce{OH^-}$ from $\ce{Ba(OH)_2}$: $n(\ce{OH^-}) = 2 \times n(\ce{Ba(OH)_2}) = 2 \times 0.06 = 0.12$ mol.

Moles of $\ce{HNO_2}$: 0.23 mol.

After reaction:

Moles of $\ce{HNO_2}$ remaining: $0.23 - 0.12 = 0.11$ mol.

Moles of $\ce{NO_2^-}$ formed: 0.12 mol.

So we have a weak acid and its conjugate base → buffer.

For $\ce{KNO_2}$:

Moles of $\ce{NO_2^-}$: 0.24 mol (from $\ce{KNO_2}$).

Moles of $\ce{HNO_2}$: 0.23 mol.

So weak acid and conjugate base → buffer.

Wait, but the options: let's check the original problem again. The options are:

- 0.24 mol HCl

- 0.06 mol $\ce{Ba(OH)_2}$

- 0.11 mol HCl

- 0.23 mol KCl

- 0.24 mol $\ce{KNO_2}$

Wait, maybe I made a mistake with $\ce{KNO_2}$. Wait, the initial solution is $\ce{HNO_2}$, adding $\ce{KNO_2}$ (which provides $\ce{NO_2^-}$) gives a buffer (weak acid + conjugate base). So both 0.06 mol $\ce{Ba(OH)_2}$ and 0.24 mol $\ce{KNO_2}$ are correct? Wait, no, let's check the moles again.

Wait, the initial moles of $\ce{HNO_2}$: $0.23\ \text{mol}$ (1 L * 0.23 M).

For $\ce{Ba(OH)_2}$:

Reaction: $\ce{2HNO_2 + Ba(OH)_2 -> Ba(NO_2)_2 + 2H_2O}$

Moles of $\ce{Ba(OH)_2}$: 0.06 mol. So moles of $\ce{HNO_2}$ reacted: $2 \times 0.06 = 0.12$ mol. Moles of $\ce{HNO_2}$ remaining: $0.23 - 0.12 = 0.11$ mol. Moles of $\ce{Ba(NO_2)_2}$ formed: 0.06 mol, so moles of $\ce{NO_2^-}$: $2 \times 0.06 = 0.12$ mol. So we have $\ce{HNO_2}$ (0.11 mol) and $\ce{NO_2^-}$ (0.12 mol) → buffer.

For $\ce{KNO_2}$:

Moles of $\ce{NO_2^-}$: 0.24 mol (from $\ce{KNO_2}$). Moles of $\ce{HNO_2}$: 0.23 mol. So we have a weak acid and its conjugate base → buffer.

Wait, but the option 0.24 mol HCl is wrong, 0.11 mol HCl is wrong, 0.23 mol KCl is wrong. So the correct options are 0.06 mol $\ce{Ba(OH)_2}$ and 0.24 mol $\ce{KNO_2}$? Wait, but let's check the buffer capacity. Wait, maybe the question is from a source where only $\ce{Ba(OH)_2}$ and $\ce{KNO_2}$ are correct? Wait, no, let's re - evaluate:

Wait, when adding a strong base to a weak acid to make a buffer, the strong base should not completely neutralize the weak acid. The moles of strong base (in terms of $\ce{OH^-}$) should be less than the moles of weak acid.

Moles of $\ce{OH^-}$ from $\ce{Ba(OH)_2}$: $0.06\ \text{mol} \times 2 = 0.12\ \text{mol}$, moles of $\ce{HNO_2}$: $0.23\ \text{mol}$. $0.12 < 0.23$, so it partially neutralizes, forming conjugate base, so buffer.

Adding $\ce{KNO_2}$: it's a salt of the conjugate base, so adding it to the weak acid gives a buffer (weak acid + conjugate base).

Wait, but the option 0.24 mol $\ce{KNO_2}$: moles of $\ce{NO_2^-}$ is 0.24, moles of $\ce{HNO_2}$ is 0.23. So yes, buffer.

But let's check the original problem again. Maybe I misread the initial concentration: "0.23 M Nitrous acid" in 1 L, so moles of $\ce{HNO_2}$ is 0.23 mol.

So the correct options are:

- 0.06 mol $\ce{Ba(OH)_2}$

- 0.24 mol $\ce{KNO_2}$

Wait, but let's check the option 0.24 mol $\ce{KNO_2}$: the formula is $\ce{KNO_2}$, not $\ce{KNO_3}$ (the last option was 0.24 mol $\ce{KNO_2}$? Wait, the last option is "0.24 mol $\ce{KNO_2}$"? Wait, the user's image shows: "0.24 mol $\ce{KNO_2}$" (wait, no, the last option is "0.24 mol $\ce{KNO_2}$"? Wait, the original text: "0.24 mol $\ce{KNO_2}$" (yes, the last option is 0.24 mol $\ce{KNO_2}$).

So the correct answers are 0.06 mol $\ce{Ba(OH)_2}$ and 0.24 mol $\ce{KNO_2}$. But let's check again:

Wait, when adding $\ce{KNO_2}$, it's a conjugate base, so $\ce{HNO_2}$ (weak acid) and $\ce{NO_2^-}$ (conjugate base) → buffer.

When adding $\ce{Ba(OH)_2}$, it's a strong base that partially neutralizes $\ce{HNO_2}$ to form $\ce{NO_2^-}$, so we have $\ce{HNO_2}$ and $\ce{NO_2^-}$ → buffer.

The other options:

- HCl (strong acid) → no buffer.

- KCl (neutral salt) → no buffer.

So the correct options are 0.06 mol $\ce{Ba(OH)_2}$ and 0.24 mol $\ce{KNO_2}$.

Answer:

0.06 mol $\ce{Ba(OH)_2}$, 0.24 mol $\ce{KNO_2}$