Question Number 170743 by LINEU last updated on 30/May/22
![2. Uma soluca^ o tampao foi preparada misturarando 200ml NH_(3 ) 0,6 moles e 300ml de uma solucao de NH_3 Cl 0,2 moles P^(kh) =9,24 e log 2=0,3 a) Qual e o P^h desta solucao tampa^ o supondo-se um volume de 500ml? b)Qual sera o P^h depois de ser adicionado 0,2 molar de ion [H^− ] Dados n(CH_3 -CH_2 COOH)=0,02mok/l=0,02M Ka=1,3∙10^(−5) n(CH_3 −CH_2 -COONa)=0,015mol/l=0,015M V=1l log 2=0,3 log 0,11 P^h =? 1°Passo 2°Passo P^(ka) =−log Ka P^h =P^(ka) +log(([Base])/([Acido])) P^(ka) =−log 1,3∙10^(−5) ? P^h =4,89+log(([0,02])/([0,015])) P^(ka) =(−5+0,11) P^h =4,89+log1,3 P^(ka) =5−11 P^h =4,89+0,11 P^(ka) =4,49 P^h =5 3°Passo P^h =P^h +log(([Basica])/([Acida])) P^h =4,89+log(([0,02])/([0,025])) P^h =4,89+log0,8 P^h =4,89+0,096 P^h =4,986≈5](https://www.tinkutara.com/question/Q170743.png)
$$ \\ $$$$\mathrm{2}.\:\mathrm{Uma}\:\:\mathrm{soluc}\overset{ } {\mathrm{a}o}\:\:\mathrm{tampao}\:\mathrm{foi}\:\mathrm{preparada}\:\mathrm{misturarando}\:\mathrm{200}\boldsymbol{\mathrm{m}{l}}\:\mathrm{NH}_{\mathrm{3}\:} \:\:\mathrm{0},\mathrm{6}\:{moles} \\ $$$${e}\:\mathrm{300}{m}\boldsymbol{{l}}\:{de}\:{uma}\:{solucao}\:{de}\:\mathrm{NH}_{\mathrm{3}} \mathrm{C}{l}\:\:\:\mathrm{0},\mathrm{2}\:{moles}\:{P}^{\mathrm{kh}} =\mathrm{9},\mathrm{24}\:{e}\:\mathrm{log}\:\mathrm{2}=\mathrm{0},\mathrm{3} \\ $$$$\left.\mathrm{a}\right)\:\mathrm{Q}{ual}\:{e}\:{o}\:{P}^{\mathrm{h}} \:\mathrm{desta}\:\mathrm{solucao}\:\mathrm{tamp}\overset{ } {\mathrm{a}o}\:\mathrm{supondo}-{se}\:{um}\:{volume}\:{de}\:\mathrm{500}\boldsymbol{{ml}}? \\ $$$$\left.\mathrm{b}\right)\mathrm{Q}{ual}\:{sera}\:{o}\:{P}^{\mathrm{h}} \:{depois}\:{de}\:{ser}\:{adicionado}\:\mathrm{0},\mathrm{2}\:{molar}\:{de}\:{ion}\:\left[\mathrm{H}^{−} \right] \\ $$$$\mathrm{Dados} \\ $$$$\mathrm{n}\left(\mathrm{CH}_{\mathrm{3}} -\mathrm{CH}_{\mathrm{2}} \mathrm{COOH}\right)=\mathrm{0},\mathrm{02}\boldsymbol{{mok}}/\boldsymbol{{l}}=\mathrm{0},\mathrm{02M} \\ $$$${K}\mathrm{a}=\mathrm{1},\mathrm{3}\centerdot\mathrm{10}^{−\mathrm{5}} \\ $$$${n}\left(\mathrm{CH}_{\mathrm{3}} −\mathrm{CH}_{\mathrm{2}} -\mathrm{COONa}\right)=\mathrm{0},\mathrm{015}{mol}/\boldsymbol{{l}}=\mathrm{0},\mathrm{015M} \\ $$$$\mathrm{V}=\mathrm{1}\boldsymbol{{l}} \\ $$$$\mathrm{log}\:\mathrm{2}=\mathrm{0},\mathrm{3} \\ $$$$\mathrm{log}\:\mathrm{0},\mathrm{11} \\ $$$$\mathrm{P}^{\mathrm{h}} =? \\ $$$$\:\:\mathrm{1}°\boldsymbol{\mathrm{Passo}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}°\mathrm{Passo} \\ $$$$\mathrm{P}^{\mathrm{ka}} =−\mathrm{log}\:\mathrm{Ka}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{P}^{\mathrm{h}} =\mathrm{P}^{\mathrm{ka}} +\mathrm{log}\frac{\left[\mathrm{Base}\right]}{\left[\mathrm{Acido}\right]} \\ $$$$\mathrm{P}^{\mathrm{ka}} =−\mathrm{log}\:\mathrm{1},\mathrm{3}\centerdot\mathrm{10}^{−\mathrm{5}} \:\:\:\:\:?\:\:\:\:\:\:\:\:\mathrm{P}^{\mathrm{h}} =\mathrm{4},\mathrm{89}+\mathrm{log}\frac{\left[\mathrm{0},\mathrm{02}\right]}{\left[\mathrm{0},\mathrm{015}\right]} \\ $$$$\mathrm{P}^{\mathrm{ka}} =\left(−\mathrm{5}+\mathrm{0},\mathrm{11}\right)\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{P}^{\mathrm{h}} =\mathrm{4},\mathrm{89}+\mathrm{log1},\mathrm{3} \\ $$$$\mathrm{P}^{\mathrm{ka}} =\mathrm{5}−\mathrm{11}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{P}^{\mathrm{h}} =\mathrm{4},\mathrm{89}+\mathrm{0},\mathrm{11} \\ $$$$\mathrm{P}^{\mathrm{ka}} =\mathrm{4},\mathrm{49}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{P}^{\mathrm{h}} =\mathrm{5} \\ $$$$\mathrm{3}°\boldsymbol{\mathrm{Passo}} \\ $$$$\mathrm{P}^{\mathrm{h}} =\mathrm{P}^{\mathrm{h}} +\mathrm{log}\frac{\left[\mathrm{Basica}\right]}{\left[\mathrm{Acida}\right]} \\ $$$$\mathrm{P}^{\mathrm{h}} =\mathrm{4},\mathrm{89}+\mathrm{log}\frac{\left[\mathrm{0},\mathrm{02}\right]}{\left[\mathrm{0},\mathrm{025}\right]} \\ $$$$\mathrm{P}^{\mathrm{h}} =\mathrm{4},\mathrm{89}+\mathrm{log0},\mathrm{8} \\ $$$$\mathrm{P}^{\mathrm{h}} =\mathrm{4},\mathrm{89}+\mathrm{0},\mathrm{096} \\ $$$$\mathrm{P}^{\mathrm{h}} =\mathrm{4},\mathrm{986}\approx\mathrm{5} \\ $$