Respuesta :
Answer:
[tex]P_H =2.86[/tex]
[tex]c=1.4\times 10^{-4}[/tex]
Explanation:
first write the equilibrium equaion ,
[tex]C_3H_6O_3[/tex] ⇄ [tex]C_3H_5O_3^{-} +H^{+}[/tex]
assuming degree of dissociation [tex]\alpha[/tex] =1/10;
and initial concentraion of [tex]C_3H_6O_3[/tex] =c;
At equlibrium ;
concentration of [tex]C_3H_6O_3 = c-c\alpha[/tex]
[tex][C_3H_5O_3^{-} ]= c\alpha[/tex]
[tex][H^{+}] = c\alpha[/tex]
[tex]K_a = \frac{c\alpha \times c\alpha}{c-c\alpha}[/tex]
[tex]\alpha[/tex] is very small so [tex]1-\alpha[/tex] can be neglected
and equation is;
[tex]K_a = {c\alpha \times \alpha}[/tex]
[tex][H^{+}] = c\alpha[/tex] = [tex]\frac{K_a}{\alpha}[/tex]
[tex]P_H =- log[H^{+} ][/tex]
[tex]P_H =-logK_a + log\alpha[/tex]
[tex]K_a =1.38\times10^{-4}[/tex]
[tex]\alpha = \frac{1}{10}[/tex]
[tex]P_H= 3.86-1[/tex]
[tex]P_H =2.86[/tex]
composiion ;
[tex]c=\frac{1}{\alpha} \times [H^{+}][/tex]
[tex][H^{+}] =antilog(-P_H)[/tex]
[tex][H^{+} ] =0.0014[/tex]
[tex]c=0.0014\times \frac{1}{10}[/tex]
[tex]c=1.4\times 10^{-4}[/tex]
