a)
Cân bằng:
\(HClO\overset {} \leftrightarrows {H^ + } + Cl{O^ - }\)
Sau cân bằng:
\({\text{[}}HClO] = 0,1.(1 - \alpha ) = 0,1.0,85 = 0,085M;{\text{[}}{H^ + }{\text{]}} = {\text{[}}C{l^ - }{\text{]}} = 0,1.\alpha = 0,1.0,15 = 0,015M\)
b)
Cân bằng:
\(C{H_3}COOH\overset {} \leftrightarrows C{H_3}COO{^ - } + {H^ + }\)
Sau cân bằng:
\({\text{[}}C{H_3}COOH{\text{]}} = 0,2.(1 - \alpha ) = 0,2.(1 - 20\% ) = 0,2.80\% = 0,16M;{\text{[}}{H^ + }{\text{]}} = {\text{[}}C{H_3}CO{O^ - }{\text{]}} = 0,2.\alpha = 0,2.20\% = 0,04M\)
c)
Cân bằng:
\(N{H_4}Cl\xrightarrow{{}}N{H_4}^ + + C{l^ - }\)
Vì \(\alpha = 1\)
\(\to {\text{[}}N{H_4}^ + {\text{]}} = {\text{[}}C{l^ - }{\text{]}} = {C_{M{\text{ N}}{{\text{H}}_4}Cl}} = 0,25M\)
d)
Ta có: \({n_{N{H_3}}} = \frac{{4,48}}{{22,4}} = 0,2{\text{ mol}} \to {{\text{C}}_{M{\text{ N}}{{\text{H}}_3}}} = \frac{{0,2}}{{0,2}} = 1M\)
Cân bằng:
\(N{H_3} + {H_2}O\overset {} \leftrightarrows N{H_4}^ + + O{H^ - }\)
Sau cân bằng:
\({\text{[}}N{H_3}{\text{]}} = 1.(1 - \alpha ) = 1.(1 - 0,1) = 0,9M;{\text{[}}N{H_4}^ + {\text{]}} = {\text{[}}O{H^ - }{\text{]}} = 1.\alpha = 0,1.1 = 0,1M\)
