• # question_answer An acidic buffer solution has $[HA]=1.0$M and$[NaA]=1.0M$. To $(10+x)$mL of this buffer solution 9 mL of $1.0M\,HCl$ is added so that pH changes by one unit. The value of $x$ is A)  $0.1$                           B)  10 C)  $1.5$                                       D)  $1.0$

For given buffer solution pH,=pKa Now 9 mL of 1 M HCl is added to (10 + x) mL of this solution $NaA\,\,\,\,\,+\,\,\,\,\,HCl\,\xrightarrow{\,}\,\,\,HA\,\,\,\,\,+\,\,\,NaCl$ $1\times (10+x)\,$ $(1\times 9)$ $1(10+x)\,\,\,\,0$ $(10+x)-9\,\,\,0$                        $(10+x)\,+9$ $p{{H}_{1}}-p{{H}_{2}}=1$ $\log \,\frac{(10+x)-9}{(10+x)\,+9}=\frac{1}{10}$ $\frac{(10+x)\,-9}{(10+x)\,+9}=\frac{1}{10}$ $x=1\,\,ml$.