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What is the concentration of NaClin a solution if titration of 15.00 $\mathrm{mL}$ of the solution with 0.2503 $\mathrm{M} \mathrm{AgNO}_{3}$requires 20.22 $\mathrm{mL}$ of the AgNO_ solution to reach the end point?$\mathrm{AgNO}_{3}(a q)+\mathrm{NaCl}(a q) \longrightarrow \mathrm{AgCl}(s)+\mathrm{NaNO}_{3}(a q)$

0.3374$M$

00:36

Sisi G.

Chemistry 101

Chapter 4

Stoichiometry of Chemical Reactions

Chemical reactions and Stoichiometry

University of Maryland - University College

Brown University

University of Toronto

Lectures

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Chemistry is the science o…

04:42

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At the end point of a titr…

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the titillation reaction that we are considering is one mole silver nitrate reacting with one mole sodium chloride producing one mole silver chloride solid and one mole sodium nitrate. To calculate polarity of the sodium chloride solution, we need to know the moles of sodium chloride and we need to know the volume of the sodium chloride solution. In leaders, we can calculate the moles of sodium chloride using the information provided for the other reactant silver nitrate. If we know the volume of the silver nitrate solution needed to completely react with all of the sodium chloride, we can convert this volume into leaders by dividing by 1000 then go from leaders of the silver nitrate solution into moles of silver nitrate by multiplying by the polarity of the silver nitrate solution, which is 0.2503 moles per liter. Now we are in units of moles silver nitrate because the stock geometry is 1 to 1. This is also the moles of sodium chloride that were in the solution being titrate, ID with the silver nitrate. We then divide the mold sodium chloride by the volume of the sodium chloride solution in leaders. So we'll convert our mill leaders into leaders by dividing by 1000. And this gives us a concentration of 10000.3374 moller.

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