PHHCL= X  [H+]HCL= 10-X PHCH3COOH = Y  [H+] = 10-Y TA CÓ

Câu 10

pH

_HCl

= x  [H

⁺

]

_HCl

= 10-x

pHCH3COOH = y  [H

⁺

] = 10-y

Ta có: HCl  H+ + Cl-

10-x  10-x M

CH3COOH  CH3COO- + H+

100. 10-y  10-y M

M t khác [HCl] = [CH3COOH]

=> 10-x = 100. 10-y

=> x = y-2