Science & Mathematics">
Cuestionario ANALITICO
Cuestionario ANALITICO
Cuestionario ANALITICO
NaOH
[𝐵] = 0.125𝑀
pH inicial
𝑝𝐻 = −𝑙𝑜𝑔[𝐻 + ]
𝑝𝐻 = −𝑙𝑜𝑔[0.150𝑀]
𝑝𝐻 = 0.8239
pH intermedio
VAc x VB= VB x VAc
(𝑉𝑥𝑁)𝐴𝑐
= 𝑉𝐵
𝑁𝐵
(25.00𝑚𝑙𝑥0.150 𝑚𝑒𝑞⁄𝑚𝑙 )𝐴𝑐
= 𝑉𝐵
0.125 𝑚𝑒𝑞⁄𝑚𝑙
𝑉𝐵 = 30.00𝑚𝑙
𝑉𝑎𝑁𝑎 − 𝑉𝑏𝑁𝑏
[𝐻 + ] =
𝑉𝑡
(25.00𝑚𝑙𝑥0.150 𝑚𝑒𝑞⁄𝑚𝑙) − (30.00𝑚𝑙𝑥0.125 𝑚𝑒𝑞⁄𝑚𝑙)
[𝐻 + ] =
55𝑚𝑙
[𝐻 + ] = 0
(25.00𝑚𝑙𝑥0.150 𝑚𝑒𝑞⁄𝑚𝑙
[𝐻 + ] =
55𝑚𝑙
[𝐻 + ] = 0.0681
𝑝𝐻 = −𝑙𝑜𝑔[0.0681]
𝑝𝐻 = 1.1668
pH eq.
𝑝𝐾𝑎 = −𝑙𝑜𝑔(𝐾𝑎)
[𝐴𝑐]
𝑝𝐻 = 𝑝𝐾𝑎-log [𝐵]
𝑝𝐾𝑎 = −𝑙𝑜𝑔(4.0x10−6 )
𝑝𝐾𝑎 = 5.3979
[0.150𝑀]
𝑝𝐻 = 5.3979-log[0.125𝑀]
𝑝𝐻 = 5.3187
2. Un volumen de 50.0 ml de una disolución 0.085M de un ácido triprótico se
titula con NaOH (Ka1=1.0x10-4, Ka2=1.0x10-7, Ka3=3.0x10-9) 0.150 M.
a) Calcule los valores de pH para los tres puntos de equivalencia.
Ac(triprotico)
V=50 ml
[𝐴𝑐] = 0.085𝑀
NaOH
[𝑁𝑎𝑂𝐻] = 0.150𝑀
FORMULAS
𝑝𝐾𝑎 = −𝑙𝑜𝑔(𝐾𝑎)
[𝐴𝑐]
𝑝𝐻 = 𝑝𝐾𝑎-log[𝑁𝑎𝑂𝐻]
Ka1=1.0x10-4
𝑝𝐾𝑎 = −𝑙𝑜𝑔(1.0x10−4 )
𝑝𝐾𝑎 = 4.0000
[0.085𝑀]
𝑝𝐻 = 4.0000-log[0.150𝑀]
𝑝𝐻 = 4.2466
Ka2=1.0x10-7
𝑝𝐾𝑎 = −𝑙𝑜𝑔(1.0x10−7 )
𝑝𝐾𝑎 = 7.0000
[0.085𝑀]
𝑝𝐻 = 7.0000-log[0.150𝑀]
𝑝𝐻 = 7.2466
Ka3=3.0x10-9
𝑝𝐾𝑎 = −𝑙𝑜𝑔(3.0x10−9 )
𝑝𝐾𝑎 = 8.5228
[0.085𝑀]
𝑝𝐻 = 8.5228log[0.150𝑀]
𝑝𝐻 = 8.7694