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PH = PKa + log [B] / [A]

Cuando es equimolar se utilize mililitro.

PH = PKa + log mlNaOH utilizado

mlAC - mlNaOH
Para el tubo 1

Para el tubo #1:
H3BO3 [ H +] + H 2 BO 3-

Ka = [ H +] [ H 2 BO 3-] = X2

[ H 3 BO 3 ] [ H 3 BO 3 ]

PKa= - log Ka

9.24 = X2 / 0.1

X2 = (9.24) (0.1)

X = √9.24 X 0.1

X = [ H +]

PH = - log √ Ka [ H +]

Ka = Antilog – Pka

Ka = Antilog (- 2.34)
0.0046 = Ka

[ H +] = √ [ 0.025 X 0.004]

[ H +] = 1.96

PH = - log [ H +]



pH=5.07



pH = pK +log [sal] = ml sol. NaOH = 0.5 = 0.5
[acido] ml acido - ml NaOH 5 - 0.5 4.5

pH = pK(ac) + log 0.54.5

pH = 9.14 + log 0.5
4.5

pH = 9.14 + (-0.95) = 8.19 para el tubo 2




pH = pK + log [sal] = ml sol. NaOH = 1.0 =1.0
[acido] ml acido - ml NaOH 5 - 1.0 4.0

pH = pK(ac) + log 1
4

pH = 9.14 + log 1
4

pH =9.14 + (-0.60) = 8.54 para el tubo 3





pH = pK + log [sal] = ml sol. NaOH = 1.5 = 1.5
[acido] ml acido - ml NaOH 5 - 1.53.5

pH = pK(ac) + log 1.5
3.5

pH = 9.14 + log 1.5
3.5

pH = 9.14 + (-0.37)= 8.77 para el tubo 4



pH = pK + log[sal] = ml sol. NaOH = 2.0 = 2.0
[acido] ml acido - ml NaOH 5 - 2.0 3.0

pH = pK(ac) + log 2.03.0

pH = 9.14 + log 2.0
3.0

pH = 9.14 + (-0.18) = 8.96 para el tubo 5



pH = pK + log [sal] = ml sol. NaOH = 2.5 = 2.5...