Hello,
After reading the solution to the question, my first question is a) How is molarity of 0.033 M of 2.5 mmol of \(NH_4^+\) ions and 2.5 mmol of \(NH_3\) molecules computed? I shall ask my other questions, if any, regarding this solution in my next post.
Common ion effect question
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Dhamnekar Winod
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Common ion effect question
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ChenBeier
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Re: Common ion effect question
2,5 mmol in 75 ml = 2,5mmol/75 ml = 0,033 mol/l = 0,033 M