1.
[imath]HCOOH \rightleftharpoons H^+ HCOO^-[/imath] [imath]K_a=1,77.10^{-4}[/imath]
[imath]H_2O \rightleftharpoons H^+ + OH^-[/imath] [imath]K_W=10^{-14}[/imath]
[imath]Ka.C_a[/imath] >> [imath]K_W[/imath] => bỏ qua sự phân li của nước
[imath]HCOOH \rightleftharpoons H^+ HCOO^-[/imath]
x----------------------x-------x
[imath]K_a[/imath]=[imath]\frac{[H^+][HCOO^-]}{[HCOOH]}=[/imath][imath]\frac{x.x}{0,1-x}[/imath]=[imath]1,77.10^{-4}[/imath]
=>x=[imath]4,12.10^{-3}[/imath]
pH=[imath]-log[H^+][/imath]=2,385
2.
Giả sử lấy 1 l [imath]H_2SO_4[/imath] => V HCOOH=1 l
=>[imath][H_2SO_4][/imath]=0,5x, [HCOOH]=0,05
pH(mới)=2,385-0,385=2=>[imath][H^+][/imath]=0,01
[imath]HCOOH \rightleftharpoons H^+ HCOO^-[/imath]
[imath]H_2SO_4 \to H^+ HSO_4^-[/imath]
[imath]HSO_4^- \rightleftharpoons H^+ SO_4^{2-}[/imath]
BT proton=>[imath][H^+]=[HCOO^-]+[HSO_4^-]+[SO_4^{2-}][/imath](1)
[imath]K_{HCOOH}[/imath]=[imath]\frac{[HCOO^-][H^+]}{[HCOOH]}[/imath]=[imath]\frac{0,01.[HCOO^-]}{0,05-[HCOO^-]}[/imath]=[imath]1,77.10^{-4}[/imath]
=>[imath][HCOO^-][/imath]=[imath]8,696.10^{-4}[/imath](2)
[imath]K_{a_2}[/imath]=[imath]\frac{[H^+][SO_4^{2-}]}{[HSO_4^-]}[/imath]=[imath]\frac{0,01.[SO_4^{2-}]}{[HSO_4^-]-[SO_4^{2-}]}[/imath]=[imath]1,2.10^{-2}[/imath]
<=>0,01[imath][SO_4^{2-}][/imath]-[imath]1,2.10^{-2}[HSO_4^-]+1,2.10^{-2}[SO_4^{2-}][/imath]=0(3)
(1)(2)(3)=>[imath][SO_4^{2-}]=3,22.10^{-3}[/imath], [imath][HSO_4^-]=5,91.10^{-3}[/imath]
=>0,5x=[imath]9,13.10^{-3}[/imath]=>x=0,01826 M
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
