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7/21/2019 Chuong 13-14-15
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Ví dụ: Xác định pH của dung dịch sau:
a. Anilin có CM = 0,2 M,6 5 2
10
C H NHK 4, 2.10
−
=
( Đ/S: pH = 8,96)
b. 6,1 g/l acid benzoic,6 5
5
C H COOHK 6, 46.10−
=
( Đ/S: pH = 2,!)
c. 0,5 M NH!OH"Cl, K NH2OH = 10#$,%1
( Đ/S: pH = 6,"9)
d. NH4Cl 0,2 M, bi&' NH! có K b = 4,$5) CH!COONa 0,01M bi&'!
5
aCH COOHK 1,*.10−
=
) CH!COONa0,01M + NH4Cl 0,2 M ( Đ/S: #,96 $% 8,&8)
e. Ki '-n 45 dng dc CH!COONa 2,0.10#2M 3i 25 dng dc HCl !,0.10 #2 M) 10 l NaOH0,1 M + 10 l CH!COOH 0,2 M. Co Ka(CH!COOH = 4,$5 ( Đ/S: #,"!' #,!)
(. CH!COOH 0,1 N,0,01α =
. ( Đ/S: pH = &)
g. H2CO! 0,025M) K 1= 4,!.10#$ 3 K 2 = 4,*.10#11 3 '7n n8ng 9 c:c ion HCO !#, CO!
2# 3 H2CO! '-ongdng dc ( Đ/S: # ' H = H*+& = -"# ' *+&
2 = #,8.-"-- , H 2*+& cn 01 = ","2&&)
. 100 l KOH 0,01 M + 100 l CH!COOH 0,02 M ;a 9ó co 3o '< 10 l NaOH 0,015 M bi&' Ka CH!COOH = 4,$5. (-&.- S34 #,88)
i. 20 l KOH 1 M + *0 l CH!COOH 1 M, Ka = 4,$5(-&.-8 5S34 5 #,2
. 6,1 g/l acid benzoic có6 5
5
C H COOHK 6, 64.10−
=
+ 1 g/l NaOH ( pH = #,-
>.2 ! NH OH NH OH"Cl+
+ N&2 !
NH OH ?NH OH"Cl 1? 1=
bi&'2
$,%1
NH OHK 10−
=
(6,"9)
+ N&2 ! NH OH ?NH OH"Cl 1? 2=
bi&'2
$,%1
NH OHK 10
−
=
(!,9)
+ N&2 !
NH OH ?NH OH"Cl 2 ?1=
bi&'2
$,%1
NH OHK 10
−
=
(6,&9)l. 100 l HCOOH 0,1M +100 l NaOH 0,05 M n& ;a >i co 'i& 3o 0,001 ol acid o- 0,001 ol>i@ ( Đ/S: &,! ' &,! ' &,9&)
. CH!COOH 0,1M 3 CH!COONa 0,1M. Bi&' CH!COOH có Ka= 4,$5. + H ca dng dc A 'aD 9Ei nF '& no n& '< 3o 1 dng dc 9ó? 0,01 ol HCl) 0,01 ol NaOH.