8.
[tex]\begin{array}{l} m{\rm{dd}}HN{O_3}\,32\% = 458,3 \times 1,2 = 549,96g\\ m{\rm{dd}}\,HN{O_3}\,14\% = 324,1 \times 1,08 = 350,028g\\ mHN{O_3} = 350,028 \times 14\% + 549,96 \times 32\% = 225g\\ nHN{O_3} = \dfrac{{225}}{{63}} = \dfrac{{25}}{7}\,mol\\ V{\rm{dd}} = 458,4 + 324,1 = 782,4\,ml\\ {C_M} = \dfrac{{\dfrac{{25}}{7}}}{{0,7824}} = 4,56M \end{array}[/tex]
9.
[tex]\begin{array}{l} m{\rm{dd}}NaOH\,30\% = 1,33 \times 200 = 266g\\ nNaOH\,trong\,{\rm{dd}}30\% = \dfrac{{266 \times 30\% }}{{40}} = 1,995\,mol\\ nNaOH\,trong\,{\rm{dd}}\,5M = 0,5 \times 5 = 2,5mol\\ {C_M}NaOH = \dfrac{{1,995 + 2,5}}{{0,5 + 0,2}} = 6,42M\\ {\rm{[}}N{a^ + }{\rm{]}} = {\rm{[}}O{H^ - }{\rm{]}} = 6,42M \end{array}[/tex]
10.
[tex]\begin{array}{l} nCa{(N{O_3})_2} = 0,2 \times 0,5 = 0,1\,mol\\ nKN{O_3} = 0,3 \times 2 = 0,6\,mol\\ V{\rm{dd}} = 0,2 + 0,3 = 0,5\,l\\ nN{O_3}^ - = 0,1 \times 2 + 0,6 = 0,8\,mol\\ {\rm{[}}N{O_3}^ - {\rm{]}} = \dfrac{{0,8}}{{0,5}} = 1,6M\\ {\rm{[}}C{a^{ + 2}}{\rm{]}} = \dfrac{{0,1}}{{0,5}} = 0,2M\\ {\rm{[}}{K^ + }{\rm{]}} = \dfrac{{0,6}}{{0,5}} = 1,2M \end{array}[/tex]
18.a
20cm3=0,02L ;50cm3=0,05L
nBa(OH)2=0,5 . 0,05=0,025(mol)
2HCl + Ba(OH)2 -> BaCl2 +2H2O
0,05mol 0,025mol
=> CM = 0,05/0,02=2,5M
b.CM Ba2+ =5/14 M
CM Cl-=5/7 M
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