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.