Question

Giúp em vs ạ

Answer 1

`[H^+]=[0,02.0,2+0,05.2.0,2]/[0,2+0,2]=0,06(M)`

  `=>pH=-log(0,06)=1,22`

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`H^[+] + OH^[-] -> H_2 O`

`0,1`        `0,1`                          `(mol)`

`n_[H^+]=1.0,1=0,1(mol)`

`n_[OH^-]=0,375.0,4=0,15(mol)`

  `=>[OH^-]=[0,15-0,1]/[0,1+0,4]=0,1(M)`

  `=>[H^+]=[10^[-14]]/[0,1]=10^[-13](M)`

  `=>pH=13`