1.

Solution of 0.1 N NH_4 OH and 0.1 N NH_4 Cl "has "P^(H)9.25. ThenP^(K_b)of NH_4 OHis

Answer»

` 9.25`
` 4.75`
` 3.75`
` 8.25`

SOLUTION :`POH = PK_b + LOG ""(S)/(B) `
` 14- 9. 25 =PK_b +log "" (0.1)/( 0.1 ) rArr PK_b =4.75`


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