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### AuthorTopic: Calculating concentration of ions in solution  (Read 14324 times) !function(d,s,id){var js,fjs=d.getElementsByTagName(s)[0];if(!d.getElementById(id)){js=d.createElement(s);js.id=id;js.src="https://platform.twitter.com/widgets.js";fjs.parentNode.insertBefore(js,fjs);}}(document,"script","twitter-wjs"); (function() {var po = document.createElement("script"); po.type = "text/javascript"; po.async = true;po.src = "https://apis.google.com/js/plusone.js";var s = document.getElementsByTagName("script")[0]; s.parentNode.insertBefore(po, s);})();

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#### govibe

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##### Calculating concentration of ions in solution
« on: September 25, 2007, 02:21:19 AM »

I was given a chemical equation and know that A 100.0 mL aliquot of 0.200 M aqueous potassium hydroxide is mixed with 100.0 mL of 0.400 M aqueous magnesium nitrate.
2KOH(aq) + 1Mg(NO3)2(aq) -> 1Mg(OH)2 + 2KNO3

I am then asked to calculate the concentration of each ion remaining in solution after precipitation is complete. The ions are K+, OH-, Mg2+, NO3-. anyone have any hints on how to do this because 0M for K+ is apparently incorrect
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#### AWK

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##### Re: Calculating concentration of ions in solution
« Reply #1 on: September 25, 2007, 03:11:22 AM »

Calculate moles of KOH and Mg(NO3)2. From stoichiometry of reaction find moles of reagent in excess and moles of Mg(OH)2 and KNO3. Neglect Mg(OH)2 (precipitate) and finally calculate concentration (volume was changed).
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