Tolong saya....!!! sebanyak 200 mL asam format HCOOH 0,1 M direaksikan dengan 50 mL larutan natrium hidroksida NaOH 0,2 M. jika harga Ka=2x10⁻⁴ dan log 2=0,3. tentukanlah pH larutan tsb.... a. 2,7 b. 3,7 c. 4,3 d.4,7 e. 5,3 terimakasih....!!!
ratih908
N HCOOH = 200x 0,1 = 20 mmol n NaOH = 50 x 0,2 = 10 mmol
HCOOH + NaOH --> NaCOOH + H2O M : 20 10 - - R : 10 10 10 10 --------------------------------------------------------------------------- S : 10 - 10 10
[H+] = Ka x asam : garam = 2 x 10-4 x 10:10 = 2x 10-4 pH = 4 - log 2 = 4-0,3 = 3,7
n NaOH = 50 x 0,2 = 10 mmol
HCOOH + NaOH --> NaCOOH + H2O
M : 20 10 - -
R : 10 10 10 10
---------------------------------------------------------------------------
S : 10 - 10 10
[H+] = Ka x asam : garam = 2 x 10-4 x 10:10 = 2x 10-4
pH = 4 - log 2 = 4-0,3 = 3,7