tinh ti le dung dich

G

giaosu_fanting_thientai

H2SO4 ==>2H+ + SO4 2-
0,02V1 0,04V1
NaOH ==> Na+ +OH-
0,035V2 0,035V2
[TEX]\frac{0,02v1 - 0,035v2}{v1+v2}=o,01[/TEX]
==>v1/v2=4,5
 
O

oc_no1_kute

dancayngheo said:
/:)/:)/:)/:)/:)/:)/:)/:)/:)/:)/:)/:)/:)/:)/:)/:)/:)/:)/:)/:)/:)/:)/:)/:)/:)/:)/:)/:)/:)/:)/:)cho dung dich x chua h2so4 0.02m va dung dich y chua naoh 0.035m .chon hai dung dich dc dung dinh z co v=v1+v2 va ph =2. tinh ti le dung dich x va dung dich y da dung
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[TEX]H_2[/TEX][TEX]S0_4[/TEX] \Rightarrow [TEX]2H^+[/TEX] +[TEX](SO_4)^2-[/TEX]
\Rightarrow [TEX]n_H_2(SO_4)[/TEX] = [TEX]2N_H^+[/TEX] = 2.0,02X mol = 0,04 X mol
[TEX]NAOH[/TEX]\Rightarrow [TEX]NA^+[/TEX] + [TEX]OH^-[/TEX]
\Rightarrow [TEX]n_NAOH[/TEX] = [TEX]n_OH^-[/TEX]= 0,035 Y mol
Ta lại có ph=2 \Rightarrow [TEX][H^+][/TEX]= [TEX]10^-2[/TEX] M
\Rightarrow [TEX][OH^-][/TEX] = [TEX]10^-12[/TEX] M
ph=2 \Rightarrow đây là môi trường axit \Rightarrow [TEX]H^+[/TEX] dư.
\Rightarrow[TEX][H^+][/TEX] =[TEX]/frac{0,04X- 0,035Y}{/frac{X+Y}[/TEX] = [TEX]10^-2[/TEX]
\Rightarrow [TEX]/frac{X}{Y}[/TEX] =5/6 :)>-:)>- ok men rôi` nhak.
 
L

lightning.shilf_bt

ban oi ! ta co the lam theo phuong phap so do duong cheo nhu sau
C(M)H2SO4 = 0,02suy ra [H+] = 0,02.2=0.04 (M)
C(M)NaOH = 0.035 suy ra [OH-] =0.035(M)
dung divh thu duoc co pH= 2 --> [H+]=-log2= 0.01(M)
lap so do duong cheo va cuoi cung ta co ti le:
x/y = (0.035-0.01)/(0.04-0.01) =0.025/0.03 =5/6
 
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