Common ion effect question

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Dhamnekar Winod
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Common ion effect question

Post by Dhamnekar Winod »

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.
molequipH1.png (126.87 KiB) Viewed 179 times
molequipH2.png (241.45 KiB) Viewed 179 times
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Re: Common ion effect question

Post by ChenBeier »

2,5 mmol in 75 ml = 2,5mmol/75 ml = 0,033 mol/l = 0,033 M
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