How many grams of NaCl are required to precipitate most of the Ag^+ ions from 2.50x10^2 mL of 0.0113 M AgNO3 solution? Write the net ionic equation for the reaction. Idk how to set this problem up over all!

Question

How many grams of NaCl are required to precipitate most of the Ag^+  ions from 2.50x10^2 mL of 0.0113 M AgNO3 solution? Write the net ionic equation for the reaction. Idk how to set this problem up over all!

anonymous · Answer

Did they give you a chemical equation with it??

xishem · Answer

Well, the precipitation reaction that will occur is going to be...

$$NaCl(aq)+AgNO_3(aq) \rightarrow AgCl(s)+NaNO_3(aq)$$Based on solubility rules, we know that AgCl will be (mostly) insoluble in water, so we can predict the reaction in that way.

Now, we know that when NaCl and AgNO_3 dissociate, they will yield the following ions...$$Na^+(aq)+Cl^-(aq)+Ag^+(aq)+NO_3^-(aq)$$To fully precipitate all of the silver(I), we need a 1:1 mole ratio of silver(I):chloride.

Since we have a concentration and volume of AgNO_3, we can calculate how many moles of AgNO_3 we have...$$c_i=\frac{n_i}{V_{sol'n}} \rightarrow n_{AgNO_3}=(c_{AgNO_3})(V_{sol'n})$$$$n_{AgNO_3}=(0.0113M)(0.250L)=2.83*10^{-3}mol\ AgNO_3$$Each mole of silver (I) nitrate that dissociates will yield one mole of Ag^+. Based on this, we know that we have...$$2.83*10^{-3}moles\ Ag^+$$To fully precipitate this, we need an equal number of moles of Cl^-, and therefore NaCl. Then, finally we need to convert to mass of NaCl...$$2.83*10^{-3}mol\ NaCl*\frac{58.443g\ NaCl}{1mol\ NaCl}=0.165g\ NaCl$$