11Chapter -Chapter 11: Reaction in Aqueous Solutions II: Calculations [11-1] Calculations involving Molarity
1
Review! o Molarity= molar concentration= M= [x] = moles solute/ L of soln o Reaction ratio/ molar ratio The relative number of moles of reactants and products shown in the balanced chemical equation Use coefficients in front of the chemical species Also called stoichiometry Mole ratios can also be obtained from the chemical formula for a substance o EX: what is the molar concentration H3O+ in .250 M soln of H2SO4? This is a diprotic (i.e. 2step.. goes from H2SO4 to HSO4(-) to SO4(2-)) and it goes from a 1:2 molar ratio... so .250M x 2 = 0.500M of H3O+ o EX: what volume of .5M H2SO4 is required to completely react with 100mL of 1M NaOH H2SO4 + 2NaOH -> Na2SO4 +2H2O.... see notebook for explanation; end result is 100mL..... find mols of NaOH, molar ratio, then molarity o Ex: 100mL of 1M NaOH and 100mL of 1M H2SO4 mixed. Determine all concentrations See notebook for work.... determine limiting reagent (no limiting reagent in this rxn); determine amount of product formed (use molar ratio); then use initial/change/end (ICE) chart.. and use molar ratio to add and subtract from products and reactants to determine the number of moles. Using these moles, divide by the volume of soln in L and determine molarity. Overview Steps: 1) write balanced equation 2) use volume and molarity to determine moles of reactants 3) use the reaction ratio to determine moles of the product 4) add the volumes of the initial solns to determine the final volume 5) molarity= mols/L o Dr. Shepler's Basic Step Write rxn Figure out how many mols (concentration X volume) Find total mols of new product (LR) Then divide mols of product by total volume M=(mols product)/(total volume) And determine the concentration of the reactant that's left over Use LR rxn to figure this out Then do (final-initial) to determine remaining amt of mols, then divide by the total volume [11-2] Titrations ON YOUR OWN [11-3] Calculations for A/B Titrations
-Chapter 11-
2
Primary standards o KHP (potassium hydrogen phthalate) Used for standardizing bases Ex: phthalic acid + KOH KHP + H2O (hydrolysis rxn) ***molar mass of KHP is 204.22 Good b/c standard it's a solid that doesn't absorb water from the air KHP is an acidic hydrogen, which means it can react with a base such as NaOH, to neutralize it and form water EX: Calculate the molarity of a NaOH soln if 27.3mL reacts with 0.4084g of KHP NaOH +KHP H2O + NaKP o Na2CO3 (sodium carbonate) Used for standardizing acids Ex: HNO3 + Na2CO3 NaNO3 + H2CO3 [11-4]Oxidation/Reduction Reactions Oxidation is the loss of electrons o The oxidation state gets larger o Na Na+ + eo 2Cl- Cl2 + 2eReduction is the gain of electrons o The oxidation state gets smaller Cl + e- ClReview the rules= Chapter 4 Redox rxns are much more difficult to balance than many other types o These rxns are most easily balanced using the half reaction method Half reaction method o Divide the equation into two half rxns One rxn for oxidation One for reduction o Balance each half reaction by number and type of atoms Balance all element other then H and O Balance O by adding H2O Balance H by adding H+ o Balance each half rxn for charge by adding electrons to the side with greater positive charge The electrons must be on opposite sides in the two equations! Oxidation: e- are products Reduction: e- are reactants
-Chapter 11o o o
3
o
o
Multiply each half rxn by some integer so that the number of electrons lost in one reaction equals the number gained by the other reaction Add the two half rxns Simplify by canceling things that appear on both sides of the equation Check the equation to make sure there are the same number and type of atoms on each side and that the total charge on both sides Ex: tin(II) ions are oxidized to tin(IV) by bromine o Ox: Sn2+ Sn4+ + 2e (reduction agent) o Red: Br2 + 2e- 2Br (oxidizing agdent) Note the phrase in acidic solution In acidic solution, H+ is added to balance the half rxns for H H+ cannot exist in basic soln!! When balancing Redox rxns in basic soln, we have to get rid of H+!! Follow the same rules used for balancing in acidic solution At the end, add OH- to form H2O with any H+ The OH- must be added to BOTH sides of the equation You will often end up with H2O on one side and OH- on the other Ex: when Cl is bubbled into basic solution, it forms hypochlorite and chloride ions.