Chemistry, 19.11.2020 18:00 alisonlebron15
What is the ka for a 0.25 m solution of a monoprotic weak acid with a ph of 4.65
Answers: 3
= [H⁺] = 2.24 x 10⁻⁵ M![\frac{[H^+][A^-]}{[HA]}](/tpl/images/0913/2274/f1369.png)
= 2x10⁻⁹
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What is the ka for a 0.25 m solution of a monoprotic weak acid with a ph of 4.65...
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