In concentrated aqueous solutions of NaCl, the Na+ (aq) ions can be so close to the
Cl- (aq) ions that a small percentage is associated in ion pairs at any given moment. Suppose that a solution contains 1.00 mol of NaCl and that 5.57% of the formula units exist in solution as such loosely bound pairs. How many ions are there in this solution?
Check my thinking on this:
1 mol NaCl contains 6.02E23 formula units of NaCl.
5.57% or 0.0557*6.02E23=?? will be present as NaCl loosley bound pairs.
The other 94.43% will ionize which will be 0.9443*6.02E23=xx formula units NaCl to ionize. Since those ionize to Na^+ and Cl^- you will have xx Na^+ and xx Cl^-. Add those together to obtain the total number of ions.
Remember to check my thinking.
Consider the following reaction: PbCl2 + 2 NaOH -> Pb(OH)2 + 2 NaCl What would be the major species observed floating in solution after one mole of PbCl2 and 2 moles of NaOH are mixed in aqueous solution? 1. Na+ ions, Cl− ions
Consider the following reaction: PbCl2 + 2 NaOH → Pb(OH)2 + 2 NaCl What would be the major species observed floating in solution after one mole of PbCl2 and 2 moles of NaOH are mixed in aqueous solution? 1. Na+ ions, Cl− ions
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