Calcular o pH após a adição de 0,(00 ; 5,00); 15,00; 25,00; 40,00; 45,00; 49,00; 50,00; 51,(00 ; 55,00 😉 e (60,00 mL) de (NaOH 0,1000) (mol L^{-1}) na titulação de (50,00 mL) de:
ácido lático (0,1000 mol L^{-1}).

Passo 1

O enunciado pede o pH ao adicionar vários volumes de \mathrm{NaOH} 0,1000\\ \mathrm{mol\,L^{-1}} em 50,00\\ \mathrm{mL} de ácido lático 0,1000\\ \mathrm{mol\,L^{-1}}. O ácido lático tem \mathrm{p}K_a=3,86, logo K_a=10^{-3,86}=1,38\times10^{-4}.
Para resolver, vamos:
– no ponto inicial ( V_\mathrm{b}=0) usar ICE para um ácido fraco;
– para 0<V_\mathrm{b}<V_\mathrm{eq}=50,00\\ \mathrm{mL} usar a equação de Henderson–Hasselbalch;
– em V_\mathrm{b}=50,00\\ \mathrm{mL} (equivalência) usar hidrólise de base conjugada;
– para V_\mathrm{b}>50,00\\ \mathrm{mL} usar excesso de \mathrm{OH^-}.

Passo 2

Antes de adicionar qualquer \mathrm{OH^-} a concentração inicial de ácido é C=0,1000\\ \mathrm{mol\,L^{-1}}. Montamos
\mathrm{HA}\rightleftharpoons \mathrm{H^+}+\mathrm{A^-},\quad K_a=\frac{[H^+][A^-]}{[HA]}.
No equilíbrio [H^+]=x, [A^-]=x, [HA]=C-x. Logo
x=\frac{-K_a+\sqrt{K_a^2+4K_aC}}{2} =\frac{-1,38\times10^{-4}+\sqrt{(1,38\times10^{-4})^2+4\cdot1,38\times10^{-4}\cdot0,1}}{2} \approx3,64\times10^{-3}\\ \mathrm{M}.
Assim
\mathrm{pH}=-\log_{10}(3,64\times10^{-3})\approx2,44.

Passo 3

Para V_\mathrm{b}<50,00\\ \mathrm{mL} temos um sistema tampão:
n\circ de mols iniciais de HA:
n_{\mathrm{HA},0}=0,1000\cdot0,05000=5,00\times10^{-3}\,\mathrm{mol}.
n\circ de mols de \mathrm{OH^-} adicionados:
n_{\mathrm{OH^-}}=0,1000\cdot V_\mathrm{b}(\mathrm{L}).
No equilíbrio sobra
n_{\mathrm{HA}}=n_{\mathrm{HA},0}-n_{\mathrm{OH^-}},\quad n_{\mathrm{A^-}}=n_{\mathrm{OH^-}}.
Então a equação de Henderson–Hasselbalch é
\mathrm{pH}=\mathrm{p}K_a+\log\frac{n_{\mathrm{A^-}}}{n_{\mathrm{HA}}} =3,86+\log\frac{n_{\mathrm{OH^-}}}{5,00\times10^{-3}-n_{\mathrm{OH^-}}}.

Passo 4

Aplicando para cada V_\mathrm{b}<50{,}00\,\mathrm{mL}:
n_{\mathrm{OH^-}}=0,1000\times V_\mathrm{b}(\mathrm{L}).
\begin{aligned} V_\mathrm{b}&=5,00\,\mathrm{mL}\Rightarrow n_{\mathrm{OH^-}}=0,00050\\ \mathrm{mol},& \mathrm{pH}&=3,86+\log\frac{0,00050}{0,00450}=2,91,\\\ V_\mathrm{b}&=15,00\,\mathrm{mL}\Rightarrow n_{\mathrm{OH^-}}=0,00150\\ \mathrm{mol},& \mathrm{pH}&=3,86+\log\frac{0,00150}{0,00350}=3,49,\\\ V_\mathrm{b}&=25,00\,\mathrm{mL}\Rightarrow n_{\mathrm{OH^-}}=0,00250\\ \mathrm{mol},& \mathrm{pH}&=3,86+\log\frac{0,00250}{0,00250}=3,86,\\\ V_\mathrm{b}&=40,00\,\mathrm{mL}\Rightarrow n_{\mathrm{OH^-}}=0,00400\\ \mathrm{mol},& \mathrm{pH}&=3,86+\log\frac{0,00400}{0,00100}=4,46,\\\ V_\mathrm{b}&=45,00\,\mathrm{mL}\Rightarrow n_{\mathrm{OH^-}}=0,00450\\ \mathrm{mol},& \mathrm{pH}&=3,86+\log\frac{0,00450}{0,00050}=4,81,\\\ V_\mathrm{b}&=49,00\,\mathrm{mL}\Rightarrow n_{\mathrm{OH^-}}=0,00490\\ \mathrm{mol},& \mathrm{pH}&=3,86+\log\frac{0,00490}{0,00010}=5,55. \end{aligned}

Passo 5

No ponto de equivalência V_\mathrm{b}=50,00\,\mathrm{mL} todo o HA virou A ^-. O volume total é 0,10000\,\mathrm{L} e [\mathrm{A^-}]=5,00\times10^{-3}/0,10000=0,0500\,\mathrm{M}. A base conjugada hidrolisa:
\mathrm{A^-}+H_2O\rightleftharpoons HA+OH^-,\quad K_b=\frac{K_w}{K_a}=\frac{10^{-14}}{1,38\times10^{-4}}=7,25\times10^{-11}.
Aproximando [OH^-]\approx\sqrt{K_b\,[A^-]}=\sqrt{7,25\times10^{-11}\cdot0,0500}=1,90\times10^{-6}\,\mathrm{M},
\mathrm{pOH}=-\log(1,90\times10^{-6})=5,72,\quad \mathrm{pH}=14,00-5,72=8,28.

Passo 6

Para V_\mathrm{b}>50,00\,\mathrm{mL} sobra \mathrm{OH^-}:
n_{\mathrm{OH^-},\mathrm{exc}}=0,1000\,V_\mathrm{b}-5,00\times10^{-3},\quad V_t=0,05000+V_\mathrm{b}.
[\mathrm{OH^-}]=\frac{n_{\mathrm{OH^-},\mathrm{exc}}}{V_t},\quad \mathrm{pOH}=-\log[\mathrm{OH^-}],\quad \mathrm{pH}=14,00-\mathrm{pOH}.

Passo 7

Calculando para V_\mathrm{b}>50{,}00\\ \mathrm{mL}:
\begin{aligned} V_\mathrm{b}&=51,00\,\mathrm{mL}\Rightarrow n_{\mathrm{exc}}=0,00510-0,00500=1,0\times10^{-4}\\ \mathrm{mol},\\\ &\quad V_t=0,10100\\ \mathrm{L},\\ [\mathrm{OH^-}]=\tfrac{1,0\times10^{-4}}{0,10100}=9,90\times10^{-4},\\\ &\quad\mathrm{pH}=14,00-(-\log9,90\times10^{-4})=11,00,\\\ V_\mathrm{b}&=55,00\,\mathrm{mL}\Rightarrow n_{\mathrm{exc}}=0,00550-0,00500=5,0\times10^{-4}\\ \mathrm{mol},\\\ &\quad V_t=0,10500\\ \mathrm{L},\\ [\mathrm{OH^-}]=\tfrac{5,0\times10^{-4}}{0,10500}=4,76\times10^{-3},\\\ &\quad\mathrm{pH}=14,00-(-\log4,76\times10^{-3})=11,68,\\\ V_\mathrm{b}&=60,00\,\mathrm{mL}\Rightarrow n_{\mathrm{exc}}=0,00600-0,00500=1,0\times10^{-3}\\ \mathrm{mol},\\\ &\quad V_t=0,11000\\ \mathrm{L},\\ [\mathrm{OH^-}]=\tfrac{1,0\times10^{-3}}{0,11000}=9,09\times10^{-3},\\\ &\quad\mathrm{pH}=14,00-(-\log9,09\times10^{-3})=11,96. \end{aligned}

Resposta

( 0,00 mL) pH =2,44
( 5,00 mL) pH =2,91
( 15,00 mL) pH =3,49
( 25,00 mL) pH =3,86
( 40,00 mL) pH =4,46
( 45,00 mL) pH =4,81
( 49,00 mL) pH =5,55
( 50,00 mL) pH =8,28
( 51,00 mL) pH =11,00
( 55,00 mL) pH =11,68
( 60,00 mL) pH =11,96

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