CHAPTER 4 REACTIONS IN AQUEOUS SOLUTION CONCENTRATION Solute Solvent Concentration Units mass NaCl / unit volume of solution g L -1 (M, molarity) concentration in moles per litre of solution c NaCl c B mol NaCl / unit volume of solution mol / litre of solution mol L -1 mol L -1 molality m NaCl mol NaCl / kg solvent molality m B mol kg -1 / kg solvent mol kg -1 Molarity, M, or the concentration in mol -l, using SI symbols, c NaCl mol NaCl Molarity volume of solution in L so the units are mol L -1 If the formula is rearranged we can get other useful relationships: moles moles Molarity volume Volume molarity Moles volume x molarity Example 1: How many mol of NaCl do I need to make 100.0 ml of a 0.157 mol L -1 solution 100.0 ml 1000 ml L x 0.157 mol L -1 100.0 x 10-3 L x 0.157 mol L 1-1 0.0157 mol Ans. 4:1
Example 2: What mass of NaOH(s) must I weigh to make 500.0 ml of a 0.250 mol L -1 solution? 500.0 ml mol required x 0.250 mol L -1 1000 ml L 1 mol required 500.0 x 10-3 L x 0.250 mol L -1 0.125 0 mol â mass required 0.125 0 mol x 39.9971 g. mol -1 5.00 g ANS Example 3: What volume of 3.00 mol L -1 H 2 SO 4 must I measure to get 1.86 g of H 2 SO 4? mol H 2 SO 4 1.86 g 98.0794 g.mol 1 1.89 6 x 10-2 mol molarity mol V. V mol molarity 1.896 x 10 2 mol 3.00 mol L 1 6.32 x 10-3 L or 6.32 ml Example 4: In a reaction 27.68 ml of 0.1025 mol L -1 silver nitrate solution is used. What mass of silver nitrate does this contain? mol molarity V moles molarity x volume mol AgNO 3 27.68 x 10-3 L x 0.1025 mol L -1 2.837 2 x 10-3 mol â mass AgNO 3 2.837 2 x 10-3 mol x 169.87 g mol -1 0.4820 g Ans 4:2
For a reaction: What volume of 3.00 mol L -1 HCl is required to react with 10.0 g of calcium carbonate? CaCO 3 (s) + 2 HCl(aq) CaCl 2 (aq) + CO 2 (g) + H 2 O (l) 10.0 g 3.00 mol L -1 10.0 g moles of CaCO 3 present 100.09 g mol 1 0.0999 1 mol 1 mol CaCO 3 requires 2 mol of HCl so moles HCl required 2 x 0.0999 1 mol 0.199 8 mol now concentration in mol L -1 moles volume in litres so volume moles concentration in mol L 1 so volume of HCl required 0.1998 mol 3.00 mol L 1 0.0666 1 L and this is 0.0666 1 L x 1000 ml L -1 66.6 ml Ans. Dilution It is often necessary to dilute a solution to the required, lower, concentration. Concentrated sulfuric acid, H 2 SO 4 (aq), is about 18.0 mol L -1. To make the bench acid (usually 6 mol L -1 or 3 mol L -1 ) the concentrated acid must be diluted. Example 1: What volume of concentrated H 2 SO 4 must I measure and dilute to give me 15.0 L of 6.00 mol L -1? First find the number of moles of acid to be present in final solution c so H 2 SO 4 moles V moles of H 2 SO 4 required c x Volume required H 2 SO 4 6.00 mol L -1 x 15.0 L 90.0 mol volume of the 18.0 mol L -1 concentrated H 2 SO 4 solution required 90.0 mol 18.0 mol L 5.00 L Ans. 1 4:3
Example 2: The label on a bottle of concentrated hydrochloric acid indicates that it contains 6 lb or 2.7 kg of the solution which is 37.8% HCl by weight and has a density 1.192 g ml -1. What is the concentration of concentrated hydrochloric acid? Mass of the contents of the bottle in grams 2.7 kg x 1000 g kg -1 2.7 x 10 3 g Mass HCl (only 37.8% HCl) in the bottle 2.7 x 10 3 g x Moles HCl present in the bottle 27. 992 mol Now we need to know the volume: mass density volume so volume 1020.6 g 36.548 g mol 1 mass density 2.7x10 3 g â volume 1.192g.mL 1 226 5 ml 2.2 65 L 27.96 mol â concentration 2.265 L 12. 3 mol L -1 37.8 100 10 20. 6 g Ans: 12 mol L -1 in HCl 4:4
Reactions Precipitation When a solution of lead(ii) nitrate is mixed with a solution of sodium iodide a yellow precipitate is formed. The equation for the reaction is: Pb(NO 3 ) 2 (aq) + 2 NaI(aq) PbI 2 (s) + 2 NaNO 3 (aq) or more concisely Pb 2+ (aq) + 2I - (aq) PbI 2 (s) yellow When IONIC SOLIDS dissolve in water - if they do - they give solutions that contain aqueous ions. These solutions are ELECTROLYTES - they conduct electricity WEAK electrolytes are slightly ionized STRONG electrolytes are highly ionized IONIC SALTS, if they dissolve, are STRONG electrolytes. We consider ionic solids to be soluble if we can at least make a 0.1 mol L -1 solution. So, what is soluble and what is not? (See table 4.1, p. 84). SOLUBLE All nitrates are soluble! All chlorides EXCEPT AgCl, PbCl 2, Hg 2 Cl 2 [and a few unusual ones!] All salts of the Group I elements and ammonium (NH 4 + ion) are soluble! Most sulfates are soluble except PbSO 4 and Group 2 (alkaline earths) CaSO 4 down (MgSO 4. is soluble) INSOLUBLE All carbonates except those of Group 1 (alkali metals) All hydroxides OH - except Ca(OH) 2 and below in Group 2 and all of the Group 1 hydroxides. S + all sulfides except those of Group 1, 2, and NH 4 When solutions are mixed, if an insoluble salt can form, it will! 4:5
Let s look at some more equations: CuSO 4 (s) copper(ii) sulfate NaOH(s) sodium hydroxide These are both soluble and their solutions react to give a pale blue precipitate of copper(ii) hydroxide (you did this in the lab). The equation is not balanced: CuSO 4 (aq) + NaOH(aq) Cu(OH) 2 (s) + Na 2 SO 4 (aq) Must have 1. mass balance 2. charge balance but it is easily balanced CuSO 4 (aq) + 2 NaOH(aq) Cu(OH) 2 (s) + Na 2 SO 4 (aq) The equation does not tell exactly what is happening. The starting solutions are really solutions of ions CuSO 4 (aq) is Cu 2+ (aq) + SO 4 (aq) (clear blue) NaOH(aq) is Na + (aq) + OH - (aq) (water white) that behave almost independently so we could rewrite the reaction to show what really happens: Cu 2+ (aq) + SO 4 (aq) + 2 Na + (aq) + 2 OH - (aq) Cu(OH) 2(s) + 2 Na + (aq) + SO 4 (aq) Total ionic equation Now, some of the ions have done nothing - SPECTATOR IONS - and can be deleted as they are not part of the reaction. Cu 2+ (aq) + 2 OH - (aq) Cu(OH) 2 (s) Net-ionic equation Copper(II) hydroxide decomposes on heating to give copper(ii) oxide, a black solid, and water Cu(OH) 2 (s) heat ( ) d CuO(s) + H 2 O(l) The oxide reacts with dilute hydrochloric acid to give a blue solution (green if too much acid is added) of copper(ii) chloride, CuO(s) + 2 HCl(aq) CuCl 2 (aq) + H 2 O(l) (Molecular) Now we can write out an equation to show things as the really are. Those species present as dissolved ions (the dissolved ionic solids in this case) and the state of all other reactants and products are shown CuO(s) + 2 H + (aq) + 2 Cl - (aq) 2 Cu 2+ (aq) + 2 Cl - (aq) + H 2 O(l) (Total ionic) TOTAL IONIC EQUATION The total ionic equation can then be simplified by removing those species that do not change in the reaction (the SPECTATOR IONS) leaving the net ionic equation for the reaction. CuO(s) + 2 H + (aq) 2 Cu 2+ (aq) + H 2 O(l) NET IONIC EQUATION 4:6
Svante Arrhenius stated that Acids and Bases Acids are species that give H + ions in solution. Bases are species that give OH - ions in solution. This turns out to be a rather general definition Ca(OH) 2 is a base H 2 SO 3 is an acid and we would describe CaO as a basic oxide and SO 2 as an acidic oxide. A basic oxide gives an aqueous solution that is basic (alkaline) An acidic oxide gives an aqueous solution that is acidic A basic oxide reacts with and acid to give a salt. An acidic oxide reacts with a base to give a salt. An amphoteric oxide reacts with both acid and base to give salts. Al 2 O 3 (s) + 6 HNO 3 (aq) nitric acid + 2 NaOH sodium hydroxide 2 Al(NO 3 ) 3 (aq) + 3 H 2 O(l) aluminium nitrate 2 NaAlO 2 + H 2 O(l) sodium aluminate Bronsted and Lowry definition an acid is a proton H + donor a base is a proton H + acceptor A strong acid is highly ionized and therefore a strong electrolyte. HCl(aq) H + (aq) + Cl - (aq) hydrochloric, hydrobromic, hydriodic, nitric, perchloric and sulfuric acids are all strong acids. A weak acid is only slightly ionized and therefore a weak electrolyte. HF(aq) î H + (aq) + F - (aq) A strong base is highly ionized and therefore a strong electrolyte. NaOH(aq) Na + (aq) + OH - (aq) alkali and alkaline earth hydroxides A weak base is only slightly ionized and therefore a weak electrolyte. NH 3 (aq) + H 2 O(l) î NH 4+ (aq) + OH - (aq) ammonia ammonium ion 4:7
Acid-Base Reaction Neutralization Strong acid/strong base NaOH(aq) base red litmus to blue The properties of the acid/base are neutralized. + HCl(aq) acid blue litmus to red Na + (aq) + salt Cl - (aq) NEUTRAL + H 2 O(l) water Na + (aq) + OH - (aq) + H + (aq) + Cl - (aq) Na + (aq) + Cl - (aq) + H 2 O(l) OH - (aq) + H + (aq) H 2 O(l) For any strong acid-strong base reaction the net ionic equation will always be the same. Weak acids are less ionized but still react with strong bases similarly. HF(aq) d so the reaction may be considered to be with HF not H +. but the overall reaction is the same. H + (aq) + F - (aq) HF(aq) + OH - (aq) H 2 O(l) + F - (aq) Weak Bases A weak base is only slightly ionized and therefore a weak electrolyte. Amines RNH 2 R 2 NH R 3 N NH 3 (aq) + H 2 O(l) î NH 4+ (aq) + OH - (aq) ammonia Primary Secondary Tertiary ammonium ion where R is an alkyl (as well as aryl ) group... something like CH 3 or CH 2 CH 3. These behave much like ammonia and are all weak bases. CH 3 CH 2 NH 2 (aq) + H 2 O(l) î CH 3 CH 2 NH 3+ (aq) + OH - (aq) 4:8
Oxidation - Reduction (- Redox Reactions) The equations for some reactions are harder to balance. Redox reactions involve electron transfer and this must also balance if an equation is to be balanced. zinc metal reacts with hydrochloric acid: The reaction is often written as Zn(s) + 2 HCl(aq) ZnSO 4 (aq) + H 2 (g) Zn(s) + 2 H + (aq) Zn 2+ (aq) + H 2 (g) zinc metal reacts with a solution of copper(ii) sulfate: Zn(s) + CuSO 4 (aq) Cu(s) + ZnSO 4 (aq) The blue colour of the Cu 2+ (aq) is replaced by the colourless Zn 2+ (aq) ions so the initial solution is blue and the final colourless (for complete reaction). 4:9
Reduction Cu 2+ (aq) + 2 e - Cu(s) gain in electrons OXIDANT Oxidation Zn(s) Zn 2+ (aq) + 2 e - loss of electrons REDUCTANT Add them together: Cu 2+ (aq) + 2 e - + Zn(s) Cu(s) + Zn 2+ (aq) + 2 e - The equation above is balanced because the reduction exactly balances the oxidation. Copper will displace silver from a solution containing silver ions.: Cu(s) + 2 AgNO 3 (aq) Cu(NO 3 ) 2 (s) + 2 Ag(s) Cu(s) + 2 Ag + (aq) Cu 2+ (aq) + 2 Ag + (aq)(aq) Again the reduction balances the oxidation at it must. What of a reaction like Fe 2+ (aq) + MnO 4- (aq) + H + (aq) Fe 3+ (aq) + Mn 2+ (aq) + H 2 O(l) the equation is not balanced - but how to? We have to know what is oxidized and what is reduced and keep track of the electrons gained and lost. Oxidation Numbers Oxidation numbers are a simple device to keep count of the electron transfer in a change. Again we choose reference points: All pure elements have oxidation number 0 Fluorine in its compounds is always given O.N. -1 Monatomic ions have oxidation number charge on the ion Hydrogen in compounds always has O.N. +1 except in metallic hydrides such as NaH where it is -1 Oxygen in compounds has O.N. -2 except in peroxides where it is -1 In simple halides the halogen has O.N. -1 The total oxidation number is equal to the charge on the molecule, formula unit or ion. In Fe 2+ the O.N. of the Fe is +2 In CaCl 2 the O.N. of the calcium is +2 and that of the Cl -1 In Cl 2 O 7 dichlorine heptaoxide the Cl is +7 and the oxygen -2 In SO 4 sulfate ion the S is +6 and the O -2 2 Cl + (7 x -2) 0 so 2 Cl - 14 0 and 2 Cl +14 S + (4 x -2) -2 so S - 8-2 and S +6 4:10
Example: Write a balanced equation for the reaction: Fe 2+ (aq) + MnO 4- (aq) + H + (aq) Fe 3+ (aq) + Mn 2+ (aq) + H 2 O(l) First figure out what is being oxidized or reduced (if anything): Fe changes from O.N. +2 to O.N. +3 so it is OXIDIZED and one electron is lost. Fe 2+ (aq) Fe 3+ (aq) + 1 e - OXIDATION The only other change in O.N. must be with the Mn which changes to Mn 2+. But what is the starting O.N. Mn + (4 x -2) -1 so here for Mn the O.N. is +7 so the change in O.N. is -5 and the Mn is reduced. MnO 4- (aq) + 5 e - Mn 2+ (aq) REDUCTION These are reduction/oxidation half equations and the are not balanced.. lets do that now. already balanced Fe 2+ (aq) Fe 3+ (aq) + 1 e - OXIDATION Fist balance the species being oxidized or reduced MnO 4- (aq) Mn 2+ (aq) (the Mn is already balanced) REDUCTION balance the oxygen with water (the reaction is in aqueous solution) MnO 4- (aq) Mn 2+ (aq) + 4 H 2 O(l) REDUCTION then we can balance the H with H+(aq) (it is in acid solution) 8 H + (aq) + MnO 4- (aq) Mn 2+ (aq) + 4 H 2 O(l) REDUCTION finally balance the charges with electrons 8 H + (aq) + MnO 4- (aq) + 5 e - Mn 2+ (aq) + 4 H 2 O(l) REDUCTION (Note: We expected 5 e - because the change in O.N. change predicted it... a good check of our work!) and the half-equation is balanced. Now combine the two half equations, at the same time balancing the electron transfer. and add 5 x Fe 2+ (aq) Fe 3+ (aq) + 1 e - 5 Fe 2+ (aq) 5 Fe 3+ (aq) + 5 e - 8 H + (aq) + MnO 4- (aq) + 5 e - Mn 2+ (aq) + 4 H 2 O(l) tidy up 5 Fe 2+ (aq) + 8 H + (aq) + MnO 4- (aq) + 5 e - 5 Fe 3+ (aq) + 5 e - + Mn 2+ (aq) + 4 H 2 O(l) 5 Fe 2+ (aq) + 8 H + (aq) + MnO 4- (aq) 5 Fe 3+ (aq) + Mn 2+ (aq) + 4 H 2 O(l) done! 4:11
Here is another one: SO 2 (g) + Cr 2 O 7 2 (aq) + H + (aq) Cr 3+ (aq) + H 2 SO 4 (aq) + H 2 O(l) Oxidation: SO 2 (g) H 2 SO 4 (aq) sulfur goes from +4 to +6 change +2 2 H 2 O(l) + SO 2 (g) H 2 SO 4 (aq) 2 H 2 O(l) + SO 2 (g) H 2 SO 4 (aq) + 2 H + (aq) 2 H 2 O(l) + SO 2 (g) H 2 SO 4 (aq) + 2 e + 2 H + (aq) Reduction: Cr 2 O 7 (aq) Cr 3+ (aq) chromium goes from +6 to +3 a change of -3... here twice Cr 2 O 7 (aq) 2 Cr 3+ (aq) Cr 2 O 7 (aq) 2 Cr 3+ (aq)+ 7 H 2 O(l) 14 H + (aq) + Cr 2 O 7 (aq) 2 Cr 3+ (aq)+ 7 H 2 O(l) 14 H + (aq) + Cr 2 O 7 (aq) + 6 e 2 Cr 3+ (aq)+ 7 H 2 O(l) Combine 3 x 2 H 2 O(l) + SO 2 (g) H 2 SO 4 (aq) + 2 e - + 2 H + (aq) 6 H 2 O(l) + 3 SO 2 (g) 3 H 2 SO 4 (aq) + 6 e - + 6 H + (aq) 14 H + (aq) + Cr 2 O 7 (aq) + 6 e - 2 Cr 3+ (aq) + 7 H 2 O(l) 3 SO 2 (g) + 8 H + (aq) + Cr 2 O 7 (aq) 3 H 2 SO 4 (aq) + 2 Cr 3+ (aq) + H 2 O(l) Not all reactions are done in acid solution! Some are in base. Example: Zn(s) + NaNO 3 (aq) + NaOH(aq) NH 3 (g) + Na 2 Zn(OH) 4 (aq) One way to balance the equation is to first of all pretend that it was done in acid solution: Zn(s) Zn(OH) 4 (aq) oxidation 4 H 2 O(l) + Zn(s) Zn(OH) 4 (aq) 4 H 2 O(l) + Zn(s) Zn(OH) 4 (aq) + 4 H + (aq) 4 H 2 O(l) + Zn(s) Zn(OH) 4 (aq) + 4 H + (aq) + 2 e - NO 3- (aq) NH 3 (g) reduction NO 3- (aq) NH 3 (g) + 3 H 2 O(l) 9 H + (aq) + NO 3- (aq) NH 3 (g) + 3 H 2 O(l) 9 H + (aq) + NO 3- (aq) + 8 e - NH 3 (g) + 3 H 2 O(l) 4:12
Now combine the 2 equations: 4 x (4 H 2 O(l) + Zn(s) Zn(OH) 4 (aq) + 4 H + (aq) + 2 e - )... 16 H 2 O(l) + 4 Zn(s) 4 Zn(OH) 4 (aq) + 16 H + (aq) + 8 e - 9 H + (aq) + NO 3- (aq) + 8 e - NH 3 (g) + 3 H 2 O(l) 13 H 2 O(l) + 4 Zn(s) + NO 3- (aq) NH 3 (g) + 4 Zn(OH) 4 (aq) + 7 H + (aq) but this is in ACID solution. What to do? Now change the reaction to one in base: Add enough OH- to both sides to neutralize the acid. 7 OH - (aq) + 13 H 2O(l) + 4 Zn(s) + NO 3- (aq) NH 3(g) + 4 Zn(OH) 4 (aq) + 7 H + (aq) + 7 OH - (aq) 7 OH - (aq) + 13 H 2O(l) + 4 Zn(s) + NO 3- (aq) NH 3(g) + 4 Zn(OH) 4 (aq) + 7 H + (aq) + 7 OH - (aq) 7 OH - (aq) + 13 H 2 O(l) + 4 Zn(s) + NO 3- (aq) NH 3 (g) + 4 Zn(OH) 4 (aq) + 7 H 2 O(l) Delete the excess water from both sides 7 OH - (aq) + 4 Zn(s) + 6 H 2 O(l) + NO 3- (aq) 4 Zn(OH) 4 (aq) + NH 3 (g) To give the final balanced equation in BASIC solution. It is generally easier to balance an ionic equation rather than a molecular equation. Titrations Acid - Base NaOH(aq) + HCl(aq) NaCl(aq) + H 2 O(l) OH - (aq) + H + (aq) H 2O(l) Indicator. Can use litmus for strong acid strong base but select to get indicator that changes colour at the STOICHIOMETRIC (EQUIVALENCE) POINT. The titration is carried out until the indicator changes colour - END POINT. If correct, we can calculate concentration. 4:13
Suppose we make a solution of NaOH and wish to know the concentration. It can be determined using a titration with HCl of known concentration (a STANDARD SOLUTION) Volume used 1 NaOH(aq) + Concentration unknown 32.98 ml 1 HCl(aq) 0.1025 mol L -1 25.00 ml 1 NaCl(aq) + 1 H 2 O(l) mol HCl used â Conc. NaOH Answer 25.00 ml 1000 ml L 1 x 0.1025 mol L 1 2.562 5 x 10-3 mol 2.5625 x 10 3 mol 32.98 x 10 3 L 0.07769 86 mol L -1 0.07770 mol L -1 Example: How much 0.1025 mol L -1 HCl is required to neutralize 25.00 ml of 0.0812 mol L -1 Ca(OH) 2 solution? First we must have a balanced equation Ca(OH) 2 (aq) + HCl(aq) CaCl 2 (aq) + H 2 O(l) Balanced molecular equation Ca(OH) 2 (aq) + 2 HCl(aq) CaCl 2 (aq) + 2 H 2 O(l) (use for titration calculations) Total ionic equation Ca 2+ (aq) + 2 OH - (aq) + 2 H + (aq) + 2 Cl - (aq) Ca 2+ (aq) + 2 Cl - (aq) + 2 H 2 O(l) 2 OH - (aq) + 2 H + (aq) 2 H 2 O(l) and divide by 2 Net-ionic equation OH - (aq) + H + (aq) H 2O(l) Mol Ca(OH) 2 0.0812 mol L -1 x 25.00 x 10-3 L 2.03 0 x 10-3 mol mol HCl(aq) 2 x 2.03 0 x 10-3 mol C mol V Volume of HCl required 2 x 2.030 x 10 3 mol 0.1025 mol L 1 0.0396 1 L 39.6 ml Ans. 4:14
Example (review): What mass of SiC (Carborundum) can be made in the reaction from 5.00 g of SiO 2 and 2.00 g of carbon. Limiting Reagent SiO 2 (s) (sand) 5.00 g + 3 C(s) 2.00 g t SiC(s) + 2 CO (g) Mol SiO 2 5.00 g 60.0843 g mol 1 C 2.00g 12.011g.mol 1 SiC max 0.0832 2 mol 1 x 0.0832 2 mol 0.166 5 mol 1/3 x 0.0555 0 mol mleast So C is limiting and the mass of SiC formed 0.0555 0 mol x 40.1 0 g mol -1 2.23 g Ans Standard Solution - a solution of accurately known concentration. Primary Standards Stable Compounds NaOH no good Solids good HCl(g) HCl(aq) î A good example of a stable solid is potassium hydrogen phthalate, HC 8 H 4 O 4 K, KHP A white crystalline solid available in high purity. This has one acidic hydrogen. O C C O O - Na + O H 4:15
Example 1 NaOH(aq) + 1 KHP(aq) 1 NaKP(aq) + 1 H 2 O(l) One mole NaOH(aq) of reacts with one mole of KHP(aq) A solution of NaOH is made rapidly dissolving 2.0 g of NaOH and making up to 500.00 ml in a graduated flask. 0.3532 g of KHP required 15.87 ml of the NaOH. What was (a) (b) the rough concentration mol NaOH 2.0 g 40.00 g. mol 1 0.0500 4 mol 0.05004 mol â conc 500 x 10 0.10 mol L 3 L -1 the standardized conc? 0.3532 g mol KHP 204.22 g mol 1.7295 x 10-3 mol 1 â conc. of the NaOH 1.7295x10 3 mol 16.87x10 3 L 0.1025 mol L -1 Ans. The NaOH solution must be protected from the CO 2 in the air!! Why? More Solutions If 100.0 ml of 0.100 mol L -1 BaCl 2 is mixed with 300.0 ml of 0.200 mol L -1 NaCl and the solution is made up to 500.0 ml, what is (a) concentration of Ba 2+ ion (b) concentration of Cl ion in the final solution? NaCl t Na + + Cl BaCl 2 t Ba 2+ + 2 Cl This is only one source of Ba 2+ but there are two sources of Cl. Total Ba 2+ in mixture 100.0 x 10-3 L x 0.100 mol L -1 1.000 x 10-2 mol 4:16
This is the only Ba 2+ the final concentration so [Ba 2+ ] 1.000 x 10 2 mol 0.5000 L Total Cl from BaCl 2 2 x 1.00 0 x 10-2 mol from NaCl 300.0 x 10-3 L x 0.200 mol L -1 2.00 x 10-2 mol L -1 Ans. Total 2.00 0 x 10-2 mol 6.00 0 x 10-2 mol 8.00 0 x 10-2 mol â c Cl or [Cl ] 8.000 x 10 2 mol 0.5000 L 0.160 mol L -1 Ans. Remember to check the sources of ions and that when the solutions are mixed the volume changes. 4:17