Consider that a single solution that is listed as 1M NaCl has a total ion concentration...

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Chemistry

Consider that a single solution that is listed as 1M NaCl has atotal ion concentration that is twice as large due to the chemicalformula containing two ions that are dissociatedin solution:

(1 mol Na+ ions + 1 mol Cl- ions)/(1 Lsoln) = 2 mol ions/L soln = 2M ions

A. If one adds 1.0L of a 1.0M NaCl solution to 1.0L of a 1.0MNaCl solution, then what is the concentration of NaCl? What is thetotal ion concentration?

B. Generalize from the previous question: if one mixes anyunreactive solutions that start with the same [ions], then whathappens to the [ions] in the mixture?

C. Extrapolate from the previous question: if one mixes twosolutions that initially have the same concentrations of ions butsome of the ions react to form an insoluble solid, then whathappens to the concentration of ions in the mixture?

D. Starting with a 10.00mL solution that is 0.300 MC5H10O5, determine theconcentration of sugar at four stages of mixing with a 7.00 MC5H10O5​. Determine theconcentration of sugar after a total of 0.50mL, 5.00mL, 500mL andeventually 5.000L of 7.00 MC5H10O5​ are added. Sketch a plotof final concentration vs. volume of the 7.00M soln. Show anyasymptotes with a dashed line.

E. If one mixes two unreactive solutions together, but thesolutions have different concentrations, then what are the possibletrends in solute concentration as one adds more and more of onesolution? As more of one is added, is there an upper or lowerlimit? Consisder all cases.

Answer & Explanation Solved by verified expert
3.9 Ratings (710 Votes)
A Total number of moles of NaCl added M1V1 M2V2 11 11 2 mol Total volume 11 2 L Concentration of NaCl number of moles volume 22 1 M Concentration of ion twice    See Answer
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