Respuesta :
Answer:
a) 6.12
b) 1.87
Explanation:
At the onset of the equivalence point (i.e the first equivalence point); alaninate is being converted to alanine.
[tex]H_2NC_2H_5CO^-_2[/tex] [tex]+[/tex] [tex]H^+[/tex] ------> [tex]H_3}^+NC_2H_5CO^-_2[/tex]
1 mole of alaninate react with 1 mole of acid to give 1 mole of alanine;
therefore 50.0 mL of 0.160 M alaninate required 50.0 mL of 0.160M HCl to reach the first equivalence point.
The concentration of alanine can be gotten via the following process as shown below;
[tex][H_3}^+NC_2H_5CO^-_2][/tex] = [tex]\frac{initial moles of alaninate}{total volume}[/tex]
[tex][H_3}^+NC_2H_5CO^-_2][/tex] = [tex]\frac{(50.0mL)*(0.160M)}{(50.0mL+50.0mL)}[/tex]
[tex][H_3}^+NC_2H_5CO^-_2][/tex] = [tex]\frac{8}{100mL}[/tex]
[tex][H_3}^+NC_2H_5CO^-_2][/tex] = 0.08 M
Alanine serves as an intermediary form, however the concentration of [tex]H^+[/tex] and the pH can be determined as follows;
[tex][H^+][/tex] = [tex]\sqrt{\frac{K_{a1}K_{a2}{[H_3}^+NC_2H_5CO^-_2]+K_{a1}K_w}{ K_{a1}{[H_3}^+NC_2H_5CO^-_2] } }[/tex]
[tex][H^+][/tex] = [tex]\sqrt{\frac{ (10^{-pK_{a1})}(10^{-pK_{a2})}(0.08)+(10^{-pK_{a1})}(1.0*10^{-14})} {(10^{-pK_{a1}})+(0.08)} }[/tex]
[tex][H^+][/tex] = [tex]\sqrt{\frac{ (10^{-2.344})(10^{-9.868})(0.08)+(10^{-2.344})(1.0*10^{-14})} {(10^{-2.344})+(0.08)} }[/tex]
[tex][H^+][/tex] = [tex]7.63*10^{-7}M[/tex]
pH = - log [tex][H^+][/tex]
pH = [tex]-log[7.63*10^{-7}][/tex]
pH= 6.12
Therefore, the pH of the first equivalent point = 6.12
b) At the second equivalence point; all alaninate is converted into protonated alanine.
[tex]H_2NC_2H_5CO^-_2[/tex] [tex]+[/tex] [tex]H^+[/tex] -----> [tex]H^+_3NC_2H_5CO^-_2[/tex]
[tex]H^+_3NC_2H_5CO^-_2[/tex] [tex]+[/tex] [tex]H^+[/tex] -----> [tex]H^+_3NC_2H_5CO_2H[/tex]
Here; we have a situation where 1 mole of alaninate react with 2 moles of acid to give 1 mole of protonated alanine;
Moreover, 50.0 mL of 0.160 M alaninate is needed to produce 100.0mL of 0.160 M HCl in order to achieve the second equivalence point.
Thus, the concentration of protonated alanine can be determined as:
[tex][H^+_3NC_2H_5CO_2H][/tex] = [tex]\frac{initial moles of alaninate}{total volume}[/tex]
[tex][H^+_3NC_2H_5CO_2H][/tex] = [tex]\frac{(50.0mL)*(0.160M)}{(50.0mL+100.0mL)}[/tex]
[tex][H^+_3NC_2H_5CO_2H][/tex] = [tex]\frac{8}{150}[/tex]
[tex][H^+_3NC_2H_5CO_2H][/tex] = 0.053 M
The pH at the second equivalence point can be calculated via the dissociation of protonated alanine at equilibrium which is represented as:
[tex]H^+_3NC_2H_5CO_2H[/tex] ⇄ [tex]H^+_3NC_2H_5CO^-_2[/tex] [tex]+[/tex] [tex]H^+[/tex]
(0.053 - x) x x
[tex]K_{a1}[/tex] = [tex]\frac{[H^+] [H^+_3NC_2H_5CO^-_2]}{[H^+_3NC_2H_5CO_2H]}[/tex]
[tex]10^{-PK_{a1}}[/tex] = [tex]\frac{x*x}{(0.053-x)}[/tex]
[tex]10^{-2.344} =\frac{x^2}{(0.053-x)}[/tex]
[tex]0.00453 = \frac{x^2}{(0.053-x)}[/tex]
[tex]0.00453(0.053-x) =x^2[/tex]
[tex]x^2+0.00453x-(2.4009*10^{-4})[/tex]
Using quadratic equation formula;
[tex]\frac{-b+/-\sqrt{b^2-4ac} }{2a}[/tex]
we have:
[tex]\frac{-0.00453+\sqrt{(0.00453)^2-4(1)(-2.4009*10^{-4})} }{2(1)}[/tex] OR [tex]\frac{-0.00453-\sqrt{(0.00453)^2-4(1)(-2.4009*10^{-4})} }{2(1)}[/tex]
= 0.0134 OR -0.0179
So; we go by the positive integer which says
x = 0.0134
So [tex][H^+]=[H_3^+NC_2H_5CO^-_2]= 0.0134 M[/tex]
pH = [tex]-log[H^+][/tex]
pH = [tex]-log[0.0134][/tex]
pH = 1.87
Thus, the pH of the second equivalent point = 1.87
