When 100 mL of 0.1 M NaCN solution is titra | vedclass

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(C) The titration involves a weak base salt $(NaCN)$ being titrated with a strong acid $(HCl)$.The reaction is: $NaCN + HCl \rightarrow HCN + NaCl$.$1

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(C) The titration involves a weak base salt $(NaCN)$ being titrated with a strong acid $(HCl)$.The reaction is: $NaCN + HCl \rightarrow HCN + NaCl$.$1$. Initially,the solution contains $NaCN$,which is a salt of a weak acid $(HCN)$ and a strong base $(NaOH)$,making the solution basic (high $pH$).$2$. As $HCl$ is added,$CN^-$ ions are converted to $HCN$. Since $HCN$ is a weak acid,a buffer solution $(HCN/CN^-)$ is formed,causing the $pH$ to decrease gradually.$3$. Near the equivalence point,the buffer capacity is exhausted,and the $pH$ drops rapidly.$4$. After the equivalence point,the solution contains excess $HCl$,causing the $pH$ to decrease further but more slowly.This behavior is represented by a curve that starts at a high $pH$ and decreases as $HCl$ is added,showing a characteristic buffer region. Graph $C$ correctly depicts this titration curve.

When $100 \ mL$ of $0.1 \ M$ $NaCN$ solution is titrated with $0.1 \ M$ $HCl$ solution,the variation of $pH$ of the solution with the volume of $HCl$ added will be:

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