Pharmacy Course General Inorganic Chemistry EXAM

Transcription

1 Pharmacy Course General Inorganic Chemistry EXAM SLUTINS 1. a) Write the name of the following compounds ( acid, hydroxide, salt, oxide, anhydride) and write out the bonds type the chemical structure and the geometry of atoms in the molecule : Ca S 2 K 2 LiH H 2 S 4 H 3 P 4 HN 2 Ca calcium oxide the elect-conf. is 3s13p1 thus the bonds with oxygen should be 1 sigma and one pigrec we can write Ca = VSEPR It a biatomic molecule thus it is linear S 2 the compounds with oxygen of sulfur which give acid are sulfurous anhydride and sulfuric anhydride ( the difference is a lone pair involved in a dative bond in the second respectively) in our case must conserve free the lone pair of its s2p2p1p1 and would link S with a dative bond and use the p1 p1 to form sigma and pigrec with the second oxygen. Thus : S VSEPR In this case 1 D 1d 1 sigma hybridization is sp2 trigonal planar K 2 In this case K belongs to the first group and then is mono valent, xygen is divalent s2p2p1p1, It is and Potassium XYDE with two sigma bonds VSEPR has two d, no D, two sigma the hybridization is sp3 Tetrahedral

2 K K LiH Lithium is of the first group, monovalent xygen is linked to hydrogen. It is an Hydroxide, Li is 2s1 and s2p2p1p1, thus we have two sigma bonds for VSEPR 2d 2sigma n=4 sp3 tetrahedral H 2 S 4 sulfuric acid sulphur has s2p2p1p1 the two hydrogens must be linked to thus two link other two oxygens we need two doublets -H S -H VSEPR 2D 0d 2 sigma n=4 sp3 tetrahedral H 3 P 4 phosphoric acid P has s2p1p1p1 thus will link three hydrogens via remains one to be linked by the unique lone pair by a dative bond -H

3 P - H -H The VSEPR gives 1D 0d three sigma, n=4 sp3 tetrahedral HN 2 nitrous acid N is s2p1p1p1 and we must leave a doublet free for the nitric acid dative bond. Thus we have to link one to hydrogen also with a sigma and one to N with one sigma and one pigrec. : N - H VSEPR 1d 0D 2sigma N=3 sp2 Trigonal planar 2. Write the rough formula of the following compounds and indicate their bonds and molecular geometry Chlorous acid Chloridric acid Perchloric anhydride Calcium bromide Lithium sulfide Magnesium perchlorate Nitric anhydride Chlorous acid is higher than hypochlorous Thus will leave intact the three doublets of chlorine thus is H--Cl for the Cl for the electron configuration we have VSEPR 3d 0D 1 sigma thus it is sp3 tetrahedral Chloridric acid idric means that it is a binary acid between hydrogen ( monovalent) and chlorine which is s2p2p2p1 thus able to form a unique sigmsa bond

4 H-Cl linear Perchloric anhydride anhydride means that contains no hydrogen, thus we must have bond between two Cl and and to be perchloric all the doublets should be saturetaed by dative bonds to oxygens. Thus the formula is Cl Cl - -- VSEPR chlorine has 3D 0d 1sigma n=4 sp3 tetrahedral Calcium bromide it is a salt by calcium hydroxide and hydrogen bromide, calcium is divalent and it is s1p1 whilst bromide is s2p2p2p1, thus the formula should be Br -- Ca----Br and we should have an VSEPR of 2sigma n=2 sp linear Lithium sulfide it is a salt by lithium hydroxide and hydrogen sulphide ( or sulfidric acid ), lithium is s1 monovalent whilst sulfide is s2p2p2p1, thus the formula should be Li-- S----Li and we should have an VSEPR of 2sigma 2d n=4 sp3 tetrahedral

5 Magnesium perchlorate The structural formula is similar to that of perchloric acid, Magnesium is s1p1, then divalent gepmetry of perchlorate is identical to that of anhydride. ( Cl ) 2 Mg Nitric anhydride we must combine N and leaveng free the doublets ( for the link of oxygen of nitric form) Thus N N VSEPR For N n = 1d, 0D, 2sigma n=3 trigonal planar 3. a) Calculate the ph of a solution of 500,0 ml containing Sodium acetate (CH 3 CNa) M? (Ka of Acetic Acid, is 1.81*10-5 M)

6 b) Calculate the ph if to the same volume of the solution in exercise a) is added moles of acetic acid, Ka of Acetic Acid, is 1.81*10-5 M) with no change in volume? a) if I know the molarity I do not need to know the volume for the calculation The compound is a salt of a weak acid and a strong basis. Thus base of the weak acid is a strong base and hydrolyzes producing H-, the ion from strong base is a weak acid and then do NT hydrolyze. Thus the formula is (H-) = SQRT (Ca Kw/ Ka) then = SQRT (( 1.00 exp-14/ 1.81 exp-5 )* ) = = 3.52 * 10-6, ph = 5.45 and then ph = 8.55 b) The procedure is similar but we added moles in L od solution, thus we have a salt ( sodium acetate 0.224M) and the acid (weak) with a concentration of / = M. Thus we are in the presence of a buffer and the formula now is (H3+ ) = Ka Ca/Cs or (H3+ )= 1.81*10-5* / = 1.81*10-5 the ph is now = Calculate the ph of a solution of HCl ( strong acid) at the concentration of: a) M b) M c) M Seems very simple the acid is completely dissociated a) ph = 3.91 b) (H3+)= 1.0 * *10-6 ph= 5.59 c) (H3+)= 1.0 * ph= 6.99

7 5. Calculate the ph of the solution containing g of potassium hydroxide (strong base) and g of acetic acid( CH 3 CH ( weak acid) ) in the same volume (500 ml) of solution. ( Ka= M)? Potassium hydroxide has a MW of = 56.00, thus we have n- moles = 0.440/ = moles, in 500 ml gives a concentration of Cb = M, For acetic acid ( MW 60.00) we have 0.355/60.00 /0.500 = There is a slight excess of base thus we have a base concentration residual Cbres = Cb-Ca, = M. Thus ph is log of = 2.40 and ph = Locate the ph of the solution of exercise 5) in a titration plot of CH 3 CH( weak acid) with KH 7. Locate the same point in the Henderson Hasselbach plot. pk 4.74

8 8. Balance the following (unbalanced) complete reaction: List the name of products. K Cl 3 + H 2 S 4 K 2 S 4 + HCl 3 The balanced reaction is 2K Cl 3 + 1H 2 S 4 1K 2 S 4 + 2HCl 3 and the products are potassium sulfate and chloric acid. Identify the limiting reagent and calculate how many grams of xx acid are formed when 6.25 grams of Potassium Chlorate and moles of sulphuric acid are put together to react. Calculation of the moles in the reaction potassium chlorate ( MW ) moles 6.25/ 92.5 = sulphuric acid ( MW =98.00) but H2S4 is given in moles if the reagents reacts 2: moles of potassium chlorate needs of sulfuric acid ( there are ) to react completely whilst potassium chlorate for the complete reaction of sulfuric acid should be moles of potassium chlorate ( there are not available, there are ) thus the limiting factor is potassium chlorate. the mole of potassium sulphate are , and the weight can be easily calculated. 9. A voltaic cell is set up using the following half-cells conditions at 25 C: Cu 2+ (aq)/cu met E 0 (Cu 2+ /Cu) = V Zn +2 (aq)// Zn Met E 0 (Zn +2 /Zn) = V

9 a) Calculate the difference of cell potential at 25 C when [Cu 2+ ] = 2.68*10-3 M, [Zn +2 ] = 2.44*10-3 M, b) Write down the half-reaction that occurs at the cathode and the one occurring at the anode and the complete balanced redox reaction and the sense of the current. c). calculate the potential of the electrochemical cell of exercise in a) if to the copper compartment is added g per liter of copper chloride ( CuCl 2 ). a) Applying the Nernst equation. The Cu cell is E=E /2 *log (2.68*10-3 M), = The Zn cell is E=E /2 *log (2.44*10-3 M), = From calculation E Cu is higher than EZn and then is the positive pole. The delta E is ( )= 1.103V The electron current goes from Zn(neg) electrode to Cu(pos). b) the reaction are Zn Zn+2 +2e Cu+2 + 2e Cu the higher potential ( positive receive electrons form the lower potential negative. The flow is from negative to positive. c) calculation of the concentration of Cu+2, number of moles 12.21g/ gives moles. These must be added to 2.68*10-3 M giving a concentration of M. Then the new ECu is E=E /2 log (0.0938)= etc. The new calculation gives a new EZn-ECu. The deltav etc. The difference of potential increases because the positive pole due to Cu is increased by the addition of copper ions to solution. 10. Describe shortly what is the difference between a complete reaction and a chemical equilibrium reaction. The complete reaction stops when one of the reagent is finished ( the limiting factor. n the contrary the equilibrium reaction stops macroscopically only when the ratio between the product of the products concentrations on the product of reagents concentrations reach the value of the equilibrium constant. Thus in the presence of products and reactants at certain values of concentrations stops. At a microscopical level the equilibrium is a continuos

10 flus od reagents in products and products in reagents leaving inaltered the ratio at equilibrium. 11. Describe shortly what is the constant of a chemical equilibrium reaction with one example. Indicate and explain the utility of the Q/K ratio. The equilibrium constant is the ratio between products and reagents which remains unaltered also if modified in one or more component. In particular if one or more reagents or products are modified, i.e. adding some component the equilibrium is removed. The system reacts going again to the equilibrium. Ti calculate if the reagents will be consumed or produced the ratio Q/K is a measure oif the distance and the verse of the reaction that the system will adopt to restore the equilibrium. Q/K >1 means that products must be reduced and the reaction goes toward left. Q/K <1 means that reagents must be reduced and the reaction goes toward right. 12. What is the difference between a covalent bond and a ionic bond. The dative bond which kind of bond is? Covalent bond require to put in common electrons by overlap of orbitals.a couple of electrons corresponds to a chemical bond. THe ionic bond occurs when none of the two atoms is so elettronegative to attract completely the electrons ( its own and that of the second atom) displacing a negative and positive charge that attract one another. Dative bond occurs when one atom is able to give a lone pair ( only as lone pair) to more electronegative atome ( in general oxygen ) to a completely void orbital. 13. Explain the role of electronegativity difference on the chemical bond.?

11 The elettronegativity is the capacity of an atom of attraction the electron density of the bond. the polarization of a bond or the ionic character of a bond is due to the difference of electronegativity between atoms bound together. Electronegativity in the Mendelejeff table increases from up to down in the column and from left to right in the rows. 14. Indicate the equilibria present in a buffer solution which give to the solution the property of buffering small additions of both acid and bases respectively. A solution containing two species ( ad example) a weak acid in presence of its salt with strong base has the caractistic of a buffer. Infact there are present two equilibria 1) the dissociation of the weak acid able to produce H3+ and 2) the hydrolysis of the base ( strong ) of the weak acid given by the salt. This second equilibrium is able to produce H-. Thus addition of small amounts of strong acid or strong bases able to drastically change ph to pure water are contrasted by this double equilibrium. In fact in firstcase ( addition of strong acid ) the buffer produces H- by hydrolysis neutralizing the effect on ph and, in the second one ( addition of strong base), the dissociation of acid produces H3+ neutralizing the effect.