QuestionJune 21, 2025

An aqueous solution contains 0.22 M sodium cyanide. 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.) 0.21 mol NaCl 0.21 mol HCN 0.11 mol Ca(OH)_(2) 0.11 mol HNO_(3) 0.22molHNO_(3)

An aqueous solution contains 0.22 M sodium cyanide. 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.) 0.21 mol NaCl 0.21 mol HCN 0.11 mol Ca(OH)_(2) 0.11 mol HNO_(3) 0.22molHNO_(3)
An aqueous solution contains 0.22 M sodium cyanide. 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.) 0.21 mol NaCl 0.21 mol HCN 0.11 mol Ca(OH)_(2) 0.11 mol HNO_(3) 0.22molHNO_(3)

Solution
4.6(296 votes)

Answer

0.21 mol HCN, 0.22 mol HNO_{3} Explanation 1. Identify Buffer Components A buffer solution consists of a weak acid and its conjugate base or a weak base and its conjugate acid. Sodium cyanide (NaCN) provides CN^-, which is the conjugate base of the weak acid HCN. 2. Evaluate Options for Buffer Formation To form a buffer with CN^-, we need to add HCN (weak acid) or an acid that can react with CN^- to form HCN. - **0.21 mol HCN**: Directly adds the weak acid needed for the buffer. - **0.22 mol HNO_{3}**: Reacts with CN^- to form HCN and NO_3^-, creating a buffer. 3. Analyze Other Options - **0.21 mol NaCl**: Does not contribute to buffer formation as it does not provide a weak acid or react with CN^-. - **0.11 mol Ca(OH)_{2}**: Adds a strong base, which does not help in forming a buffer with CN^-. - **0.11 mol HNO_{3}**: Insufficient amount to convert enough CN^- into HCN for effective buffering.

Explanation

1. Identify Buffer Components<br /> A buffer solution consists of a weak acid and its conjugate base or a weak base and its conjugate acid. Sodium cyanide ($NaCN$) provides $CN^-$, which is the conjugate base of the weak acid $HCN$.<br /><br />2. Evaluate Options for Buffer Formation<br /> To form a buffer with $CN^-$, we need to add $HCN$ (weak acid) or an acid that can react with $CN^-$ to form $HCN$. <br /><br />- **0.21 mol HCN**: Directly adds the weak acid needed for the buffer.<br />- **0.22 mol $HNO_{3}$**: Reacts with $CN^-$ to form $HCN$ and $NO_3^-$, creating a buffer.<br /><br />3. Analyze Other Options<br /> - **0.21 mol NaCl**: Does not contribute to buffer formation as it does not provide a weak acid or react with $CN^-$. <br /> - **0.11 mol $Ca(OH)_{2}$**: Adds a strong base, which does not help in forming a buffer with $CN^-$.<br /> - **0.11 mol $HNO_{3}$**: Insufficient amount to convert enough $CN^-$ into $HCN$ for effective buffering.
Click to rate:

Similar Questions