ଏକକ 3 ବିଦ୍ୟୁତ୍ ରସାୟନ (ଅନ୍ତର୍ଗତ ପ୍ରଶ୍ନ-3)

ଉତ୍ତର

ଏକ ଦ୍ରବଣର ପ୍ରବାହିତା ହେଉଛି ଦ୍ରବଣର ଏକକ ଆୟତନରେ ଉପସ୍ଥିତ ଆୟନଗୁଡ଼ିକର ପ୍ରବାହିତା। ଦ୍ରବଣକୁ ଲଘୁକରିବା ସମୟରେ ଆୟନଗୁଡ଼ିକର ସଂଖ୍ୟା (ଯାହା ବିଦ୍ୟୁତ୍ ପ୍ରବାହ ବହନ କରିବା ପାଇଁ ଦାୟୀ) ହ୍ରାସ ପାଏ। ଫଳସ୍ୱରୂପ, ଦ୍ରବଣର ପ୍ରବାହିତା ଲଘୀକରଣ ସହିତ ହ୍ରାସ ପାଏ।

ଉତ୍ତର

$ \wedge_m^{o}({H}2 {O})=\lambda{{H}^{+}}^{o}+\lambda^{o} {OH}^{-} $

$ \text {We find out ;} \quad \wedge_m^{o}({HCl}), \wedge_m^{o}({NaOH}) \text { and } \wedge_m^{o}({NaCl}) .$

$\text {Then,} \wedge_m^{o}({H}_2 {O})=\wedge_m^{o}({HCl})+\wedge_m^{o}({NaOH})-\wedge_m^{o}({NaCl})$

ଉତ୍ତର

$C=0.025 {~mol} {~L}^{-1}$

$\Lambda_{m}=46.1 {\hspace{0.5mm} S\hspace{0.5mm}cm}^{2} {~mol}^{-1}$

$\lambda^{o}\left({H}^{+}\right)=349.6 {\hspace{0.5mm} S\hspace{0.5mm}cm}^{2} {~mol}^{-1}$

$\lambda^{o}\left({HCOO}^{-}\right)=54.6 {\hspace{0.5mm} S\hspace{0.5mm}cm}^{2} {~mol}^{-1}$

$\Lambda_{m}^{o}({HCOOH})=\lambda^{o}\left({H}^{+}\right)+\lambda^{o}\left({HCOO}^{-}\right)$

$\quad\quad\quad\quad\quad\quad\quad=349.6+54.6$

$\quad\quad\quad\quad\quad\quad\quad=404.2 {\hspace{0.5mm}S\hspace{0.5mm}cm}^{2} {~mol}^{-1}$

ଏବେ, ବିଯୋଜନ ଡିଗ୍ରୀ:

$ \begin{aligned} \alpha & =\frac{\Lambda_{m}({HCOOH})}{\Lambda_{m}^{o}({HCOOH})} \\ \\ & =\frac{46.1}{404.2} \\ \\ & =0.114 \text { (ପ୍ରାୟ) } \end{aligned} $

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad{HCOOH} \rightleftharpoons {HCOO}^{-}+{H}^{+}$

$\text{Initial conc.} \quad\quad\quad\quad\quad c {~mol} {~L}^{-1}$

$\text{Conc. at eqm.} \quad\quad\quad\quad\hspace{0.1cm} c(1-\alpha) \quad\quad c \alpha \quad\quad\quad c \alpha$

ତେଣୁ, ବିଯୋଜନ ସ୍ଥିରାଙ୍କ:

$ K_a =\dfrac{c \propto^{2}}{(1-\propto)} $

$K_a =\dfrac{\left(0.025 \right)(0.114)^{2}}{(1-0.114)} $

$K_a =3.67 \times 10^{-4} $