حل تشریحی سوال شماره 19 شیمی 3

روش اول:

$\alpha =\frac{\left[H^{+}\right]}{\left[HA\right]}\to \frac{{\alpha }_{HX}}{{\alpha }_{HA}}=\frac{\frac{{\left[H^{+}\right]}_{HX}}{\left[HX\right]}}{\frac{{\left[H^{+}\right]}_{HA}}{\left[HA\right]}}\to \frac{2}{1}=\frac{\frac{{\left[H^{+}\right]}_{HX}}{0/05}}{\frac{{\left[H^{+}\right]}_{HA}}{0/1}}\to 2\cdot 0/05{\left[H^{+}\right]}_{HX}=0/1\cdot {\left[H^{+}\right]}_{HX}\to$

${\left[H^{+}\right]}_{HA}={\left[H^{+}\right]}_{HA}\to p{H}_{HA}=p{H}_{HX}$

(0/25) (0/25)

روش دوم:

$\alpha =\frac{\left[H^{+}\right]}{\left[HA\right]}\to {\left[H^{+}\right]}_{HA}=0/1{\alpha }_{HA},{\left[H^{+}\right]}_{HX}=0/5\times {\alpha }_{Hx}$

${\to }_{}^{2_{Hx}=2{\alpha }_{HA}}\to {\left[H^{+}\right]}_{HX}=0/05\times 2{\alpha }_{HA}=0/1{\alpha }_{HA}\to [H^{+}{]}_{H}A=[H^{+}{]}_{H}x\to p{H}_{HA}=p{H}_{Hx}$

(0/25) (0/25) (0/25)