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.
Your thinking is mostly correct. Let's go through the calculations step by step to find the total number of ions in the solution.
Step 1:
You are correct that 1 mole of NaCl contains 6.02 x 10^23 formula units of NaCl.
Step 2:
To find the number of formula units present as loosely bound pairs, you need to calculate 5.57% (0.0557) of the total number of formula units in 1 mole of NaCl. So, multiply 0.0557 by 6.02 x 10^23:
0.0557 x 6.02 x 10^23 = 3.34 x 10^22 formula units of NaCl in loosely bound pairs.
Step 3:
To find the number of formula units that will ionize, subtract the number of formula units in loosely bound pairs from the total number of formula units in 1 mole of NaCl:
6.02 x 10^23 - 3.34 x 10^22 = 5.69 x 10^22 formula units of NaCl to ionize.
Step 4:
Since each NaCl formula unit ionizes into one Na+ ion and one Cl- ion, the number of Na+ ions and Cl- ions will be the same as the number of formula units to ionize:
Number of Na+ ions = Number of Cl- ions = 5.69 x 10^22 ions.
Step 5:
To find the total number of ions in the solution, add the number of Na+ ions and Cl- ions together:
Total number of ions = Number of Na+ ions + Number of Cl- ions = 5.69 x 10^22 + 5.69 x 10^22 = 1.138 x 10^23 ions.
Therefore, there are 1.138 x 10^23 ions in the solution.