Download Presentation
## Solution Chemistry

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -

**Solution Chemistry**Dealing with mixtures**Solutions**A solution is a homogenous mixture consisting of a solvent and at least one solute. The solvent is the most prevalent species. The solute is the less prevalent species.**Examples of Solutions**Saline (salt water) is a solution. The solvent is water, the solute is salt. Wet salt is also a solution. The solvent is salt, the solute is water.**160 proof Vodka**What is the solvent? Alcohol – It is 80% alcohol. What is the solute? Water – It is 20% water.**Aqueous Solutions**Aqueous solutions are specifically solutions where water is the solvent. Aqueous solutions are a very common medium for performing chemical reactions.**Advantages of Aqueous Solutions**• Mixing – you can stir the solution. • Ability to dissipate heat (or cold) – the mass of the solvent allows it to absorb significant amounts of heat (or cold). • “Universal solvent” – water dissolves many different materials, especially ionic materials.**Concentration**Because a solution is a mixture – there are different ratios of solvent/solute quantities possible. For example, I could put 1 teaspoon of salt in a cup of water OR I could put 2 teaspoons of salt in a cup of water. Both are saline solutions, but they have different amounts of salt.**Concentration**Almost any unit of measure can be used to specify concentration. (teaspoon solute/cup solvent would work!) There are certain common units of measuring solution concentration that are most frequently used. Understanding their UNITS! UNITS! UNITS! And being able to manipulate those UNITS! UNITS! UNITS! is crucial.**Common units of concentration**% by mass – % by volume % by mass-volume Molarity – Molality – Normality ppt – ppm – ppb – lb/million gallons -**Converting units**What is the molarity of a 10% by mass aqueous NaCl solution? UNITS! UNITS! UNITS! 10 g NaCl = moles NaCl 100 g NaCl solution L solution To convert g NaCl to moles, you need to know… Molar mass of NaCl To convert g solution to L solution, you need to know… Density of the solution**The Density**We ALWAYS know the molar mass of any substance. But what about the density?**The Density**You don’t always know the density. Density depends on concentration. Sometimes you know the density. Sometimes you can figure out the density. Sometimes you just have to ASSUME the density.**The Density**If you don’t know anything except what was given: What is the molarity of a 10% by mass aqueous NaCl solution? What would you do? Assume the density is that of pure water (1.0 g/mL at 25°C)**The Density**Suppose I had further information: What is the molarity of a 10% by mass aqueous NaCl solution? (Density of 5% NaCl solution = 1.05 g/mL, Density of 20% NaCl solution = 1.13 g/mL) Now what would you do? I can either ASSUME that 5% is “close enough” to 10%. OR I can “interpolate” the density between 5% and 20%.**Linear Interpolation**Do you know what a “linear interpolation is”? I assume that there is a linear (straight-line) dependence of the density on the concentration. (By the way, this is not true, but it is an OK assumption if the range is narrow enough.) Then I draw a straight line between the two points I know and find the interpolated concentration at my concentration of interest.**You can also do it without the graph.**What is the molarity of a 10% by mass aqueous NaCl solution? (Density of 5% NaCl solution = 1.05 g/mL, Density of 20% NaCl solution = 1.13 g/mL) I find the slope of the line: Δ Density Δ % NaCl 1.13 g/mL – 1.05 g/mL = 5.33x10-3g/mL 20% - 5 % %**My Problem**What is the molarity of a 10% by mass aqueous NaCl solution? (Density of 5% NaCl solution = 1.05 g/mL, Density of 20% NaCl solution = 1.13 g/mL) 5.33x10-3g/mL means that every 1% change in concentration % results in a 5.33x10-3 g/mL change in density 10%-5% = 5% change 5.33x10-3g/mL * 5 % = 0.0267 g/mL change % 1.05 g/mL + 0.0267 g/mL = 1.077 g/mL = 1.08 g/mL interpolated density**Solving the problem**What is the molarity of a 10% by mass aqueous NaCl solution? (Density of 5% NaCl solution = 1.05 g/mL, Density of 20% NaCl solution = 1.13 g/mL) 10 g NaCl * 1 mol NaCl = 0.171 mol NaCl 100 g solution 58.45 g NaCl 100 g solution 0.171 mol NaCl * 1.08 g solution * 1000 mL = 1.84 mol NaCl 100 g solution 1 mL solution 1 L L solution**Converting units**Typically speaking, you can convert any of the concentration units into any of the others as long as you have the Molar Mass and the Density!**Density – your critical judgment**For a solution, sometimes you know the density, sometimes you don’t. There are tables, but they are not all inclusive. You might, for example, find in a table that: Density (30% HCl) = 1.12 g/mL Density (40% HCl) = 1.23 g/mL Density (36% HCl) = ???**Interpolate or Assume**Density (30% HCl) = 1.12 g/mL Density (40% HCl) = 1.23 g/mL Density (36% HCl) = ??? You could assume that 36% is closest to 40% and use 1.23 g/mL. This is legitimate, although not 100% accurate. Results may vary, depending on how good the assumption is.**Interpolate or Assume**Density (30% HCl) = 1.12 g/mL Density (40% HCl) = 1.23 g/mL Density (36% HCl) = ??? You could assume that density changes linearly with concentration (it doesn’t, but it is pseudo-linear for small changes). In that case, you would “linearly interpolate” the density. 1.23 g/mL – 1.12 g/mL = 0.011 g/mL = 0.011 g 40% HCl-30%HCl % mL% 1.12 g/mL + 0.011 g/mL% * 6% = 1.186 g/mL = 1.19 g/mL This is legitimate, although still not 100% accurate, but probably better than the previous assumption.**If I don’t have Density tables…**For dilute solutions, you can get pretty close by assuming the density of the solution is the same as the density of pure water. For concentrated solutions (like 36%), this is probably not a good assumption, but it is better than nothing!**Further Example**56.0 g of Fe2O3 was dissolved in water yielding a total solution volume of 2.65 L. What is the molarity of the resulting solution? 56.0 g Fe2O3 * 1 mol Fe2O3 = 0.351 mol Fe2O3 159.69 g Fe2O3 0.351 mol Fe2O3= 0.132 M Fe2O3 2.65 L**What’s it all about?**MOLES! MOLES! MOLES! Specifically, doing reactions!**An example**56.50 mL of a 2.15 M ammonium sulfate solution is mixed with 36.0 g of iron (III) chloride. If the reaction proceeds with a 65% yield, how much iron (III) sulfate would be acquired?**Limiting Reagent Problem**What’s the first thing you need? A balanced equation! (NH4)2SO4 + FeCl3 → Fe2(SO4)3 + NH4Cl How do you know this is the right products? Charges! This is an example of a double replacement reaction. The cations get switched (or the anions, if you prefer).**Limiting Reagent Problem**We still need to balance it! (NH4)2SO4 + FeCl3 → Fe2(SO4)3 + NH4Cl 3 (NH4)2SO4 + 2 FeCl3 → Fe2(SO4)3 + 6 NH4Cl**Armed with Stoichiometry!**56.50 mL of a 2.15 M ammonium sulfate solution is mixed with 36.0 g of iron (III) chloride. If the reaction proceeds with a 65% yield, how much iron (III) sulfate would be acquired? 3 (NH4)2SO4 + 2 FeCl3 → Fe2(SO4)3 + 6 NH4Cl**Armed with Stoichiometry!**56.50 mL of a 2.15 M ammonium sulfate solution is mixed with 36.0 g of iron (III) chloride. If the reaction proceeds with a 65% yield, how much iron (III) sulfate would be acquired? 3 (NH4)2SO4 + 2 FeCl3 → Fe2(SO4)3 + 6 NH4Cl 36.0 g FeCl3 *1 mol FeCl3 * 1 mol Fe2(SO4)3 * 399.87 g Fe2(SO4)3= 44.37 g Fe2(SO4)3 162.21 g FeCl3 2 mol FeCl3 1 mol Fe2(SO4)3**Armed with Stoichiometry!**56.50 mL of a 2.15 M ammonium sulfate solution is mixed with 36.0 g of iron (III) chloride. If the reaction proceeds with a 65% yield, how much iron (III) sulfate would be acquired? 3 (NH4)2SO4 + 2 FeCl3 → Fe2(SO4)3 + 6 NH4Cl Like any ratio of units, this is really just a conversion factor!!!**Armed with Stoichiometry!**56.50 mL of a 2.15 M ammonium sulfate solution is mixed with 36.0 g of iron (III) chloride. If the reaction proceeds with a 65% yield, how much iron (III) sulfate would be acquired? 3 (NH4)2SO4 + 2 FeCl3 → Fe2(SO4)3 + 6 NH4Cl**Armed with Stoichiometry!**56.50 mL of a 2.15 M ammonium sulfate solution is mixed with 36.0 g of iron (III) chloride. If the reaction proceeds with a 65% yield, how much iron (III) sulfate would be acquired? 3 (NH4)2SO4 + 2 FeCl3 → Fe2(SO4)3 + 6 NH4Cl Limiting Reagent is (NH4)2SO4: 16.13 g Fe2(SO4)3 theoretical 16.13 g Fe2(SO4)3 theoretical * 65 g actual = 10.48 g actual Fe2(SO4)3 100 g theoretical**Clicker Question**I have 1 L of a solution that is 5.4% by mass sodium sulfate. If the density of 5% sodium sulfate is 1.085 g/mL, how much silver (I) chloride would I need to add to precipitate all of the sulfate? A. 59 g B. 257 g C.118 g D. 129 g E. 25.4 g**Na2SO4 + 2 AgCl 2 NaCl + Ag2SO4**1L * 1000 mL * 1.085 g *5.4 g Na2SO4= 58.59 g Na2SO4 1L 1 mL 100 g solution 58.59 g Na2SO4 * 1 mol Na2SO4= 0.4126 mol Na2SO4 142 g Na2SO4 0.4126 mol Na2SO4 * 2 mol AgCl = 0.825 mol AgCl 1 mol Na2SO4 0.825 mol AgCl * 143 g AgCl = 118 g AgCl 1 mol AgCl**Question**When 50.00 mL of 0.125 M silver (I) nitrate is mixed with 50.00 mL of 0.250 M sodium sulfate a greyish solid forms. If I recover 0.813 g of solid, what is the yield of the reaction?