acidic basic neutral cannot determineCoCl3Na2SeBaI2BaSa) Rank the solutions in order of increasing vapor pressure. Write the formula for one solute in each box.c) Which of the following solutions has the smallest change in entropy due to the salt dissolving? Again assume complete dissolution but consider real solutions. (Circle one)0.3 m KClO4(aq)0.2 m BaI2(aq)they experience the same change in entropyGTTf pured) On the plot to the right:1) Sketch the Gibbs free energy vs. temperature for an aqueous solution of KClO4(ΔHsol= 51 kJ/mol). 2) Show the y-intercept relative to pure water. 3) Indicate and label the freezing point of the solution.ONew freezing point101046-3 per arrow-2 per mistake2 ptsΔH2 ptsslope2 pts freezing point must be below TmH2Te
CHEM230 _W16 _PREP 3 _keyp 27436179)OAsHOOHOHAsHOOHOHA)B)A is the stronger acid with the lower pKabecause the conjugate base can be resonance stabilized as show below.a) Circle the compound with the lower pKaOAsHOOOHOAsHOOOHc) Explain why isothiocyanic acid (pKa= 1.1) has a lower pKathan cyanic acid (pKa= 3.45)H-S-CNH-O-CNisothiocyanic acidpKa= 1.1cyanic acidpKa= 3.45The conjugate base of HSCN has a negative charge on the larger S atom where it is more dispersed and thus more stable whereas the HOCN has the negative chage on the smaller O atomOR HS bond is weaker and easier to break due to the greater disparity in size of H and S than H and O.b) Briefly explain your choice224) a) Write out the equilibrium that would have a K = 8x10-11given that Fe+2as Fe(H2O)6+2has a Ka= 8x10-11. Included phases for all reactants and products.b) Using one or two sentences, explain why Fe+2(Ka= 8x10-11) has a smaller Kathan Fe+3(Ka= 3.5x10-3.)8Fe(H2O)6+2(aq)+ H2O(l)<--> H3O+(aq)+ Fe(H2O)5(OH)+1(aq)The higher charge on the Fe+3better stabilizes the negative charge(on the OH-) on the conjugate base making the products more stable and thus made in larger proportion, resulting in a higher K OR gerater the charge, weaken O-H bond in water, making it easier to break the bond and donate a protont- 1 charge-2 incorrect conj. base
CHEM230 _W16 _PREP 3 _keyp 28214) 50.0 mL of 6.0 M NaOH(aq)(MW = 40) is mixed with 150.0 mL of 4.5 M NaCl(aq)(MW= 58.44).The density of the resulting solution is 1.027 g/mL. Answer the two questions about the combined solution.a) What is the pH of the combined solution?200 mL of solution0.05L6molNaOHL!"#$%&=0.3molNaOH→0.3molOH−0.3molOH−0.2Lsolution=1.5MOH−pOH=−log[OH−]=−log(1.5)=−0.176pH=14−pOH=14−(−0.176)pH=14.176pH =14.176b) If the vapor pressure of pure water at 20oC is 17.5 torr, what is the vapor pressure of the combined solution at 20oC?Pvap,solution= torr14.251 pts mol OH1 pts volume solution2 pts [OH-] (or if used M1V1 = M2V2 then 4 pts here instead of 1+1+22 pts pOH2 pts pH82 pts mass solution2 pts mass NaOH2 pts mass NaCl1 pts mass water; 1 pt mole water200mLsolution1.027gmL!