Lista de Exercícios Resolvida-2022 2

Catalise acidosa de funçao: Lucom (Ac-)- Dado: \[HNO_2] = 0,120 \ mol \cdot L^{-1}\ Ka = 7,1 \times 10^{-4}\ pH=? Resolucao HNO_2 \ (aq) + H_2O \ (l) \leftrightarrow H_3O^+ \ (aq) + NO_2^- \ (aq) Inicio \ 0,120 \ - \ - \ - Equilibrio \ 0,120-x \ - \ x \ x No equilibrio: [H_3O^+]=[NO_2^-] = x \ Ca=[HNO_2]+[NO_2^-] 0,120=[HNO_2] + x \ [HNO_2]=0,120-x Ka=\frac{[H_3O^+][NO_2^-]}{[HNO_2]} \Rightarrow \ 7,1 \times 10^{-4} = \frac{x\cdot x}{0,120-x} Resolvendo \ pelo \ metodo \ de \ aproximacao: 7,1 \times 10^{-4} = \frac{x \cdot x}{0,120} \ , \ Ca\approx0,120\approx[HNO_2] \approx \frac{x \cdot x}{0,120} Rightarrow \ x^2=7,1 \times10^{-4}\cdot0,120 \ x=\sqrt{8,52 \times 10^{-5}} \Rightarrow \ x=[H_3O^+]=9,23\times10^{-3}\ M \ pH=-\log \ [H_3O^+] pH=-\log(9,23\times10^{-3}) \Rightarrow\boxed{pH=2,03} Resolvendo \ a \ equacao \ quadratica: 7,1 \times 10^{-4} = \frac{x^2}{0,120-x} Rightarrow x^2=7,1 \times 10^{-4}\cdot(0,120-x) x^2=8,52\times10^{-5}-7,1\times10^{-4} x^2+(7,1\times10^{-4})x-8,52\times10^{-5}=0 a=1\ b=7,1\times10^{-4}\ e\ c=-8,52\times10^{-5} \Delta =b^2-4\cdot a \cdot c \Delta =(7,1\times10^{-4})^2-4\cdot1\cdot(-8,52\times10^{-5}) \Delta =5,04\times10^{-7}+3,41\times10^{-4}\ -\Delta \=3,41\times10^{-4} \Delta =3,41\times10^{-4}\Rightarrow\Delta=5,04\times Rightarrow x= +3,41\times10^{-4} 2\cdot1 \Rightarrow x^{'}=-7,1\times10^{-4}+0,0082474 \Rightarrow x^{'}=9,28\times10^{-3}\ \checkmark x^{''}=-7,1\times10^{-4}-0,0082474 \Rightarrow x^{''}=-9,19\times10^{-3} \times \ (Nao \ existe \ concentracao \ negativa) Entao: [H_3O^+]=x^{'}=9,28\times10^{-3}\ M \ pH=-\log \ [H_3O^+] pH=-\log(9,28\times10^{-3}) \Rightarrow \boxed{pH=2,03}