Asam lemah HCOOH 0,1 M dalam air terionisasi sebanyak 5% harga tetapan ionisasi asam lemah tersebut .... Question from @maeyfajrul

September 2018 2 11 Report

Asam lemah HCOOH 0,1 M dalam air terionisasi sebanyak 5% harga tetapan ionisasi asam lemah tersebut adalah.?

reniasriati81 Α = √Ka/Ma
5% = √Ka/0,1
0,05 = √Ka/0,1
25 x 10⁻⁴ = Ka/ 10⁻¹
Ka = 25 x 10⁻⁵ = 2,5 x 10⁻⁴

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maeyfajrul terima kasih kak

Risep $Ka= \alpha ^2.M \\ \\ =(5.10^{-2})^2.10^{-1} \\ \\ =25.10^{-5} \\ \\ =2,5.10^{-4}$

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maeyfajrul trimaksih kak

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