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- Consider citric acid, H3C6H5O7, added to many soft drinks. The equilibrium constants for its step-wise ionization areKa1=7.5104 ,Ka2=1.7105, andKa3=4.0107. Write the overall net ionic equation and calculate K for the complete ionization of citric acid.8. Our bodies are able to preserve a near constant pH due to the presence of hydrogen carbonate anion (HCO,). K values for the diprotic acid carbonic acid (H,CO;) are Kaj = 4.5 x 10?, K2 = 4.7 x 10-". a) Hydrogen carbonate anion is an amphoteric species. Define amphoteric. b) In the space below, write out chemical equations to describe the behavior of this species as both a Bronsted acid and a Bronsted base, and use the equilibrium constants given above to decipher whether a solution of HCO;¯ will be acidic, basic, or neutral. Support your answer with pertinent calculations.Explain why we can simplify the calculations for K when the value of K is very small (less 4.1 than -10*). 4.2 HCN will react very slightly with water with a K = 6.2 x 10-10. HCN(aq) + H,O() H,O* (aq) and CN" (aq) A chemist starts with a 0.50 M HCN solution. What is the equilibrium concentration of H;O*?. Then use the concentration of H3O* to calculate the pH of the solution with the equation pH = -log[H3O*] (you can do this!!!). Put your final answers in the box. If you need more space to show your work, you can use the backside of this page. [H;O'] at equilibrium pH = -log(H,O] pH = % of HCN that reacted
- The hydronium ion concentration of an aqueous solution of 0.558 M isoquinoline (a weak base with the formula CoH-N) is ... M. = LO°H]1L hem101 M Inbox-elaine.mirandaperez X 9 Microsoft PowerPoint- chap X G experimento sobre ejercicio X Question 10 of 11 Determine concentration of hydroxide ion, OH-, in a solution of HNO, by constructing an ICE table, writing the equilibrium constant expression, and using this information to determine the concentration of OH-. Complete Parts 1-3 before submitting your answer. NEXT 2. A 0.60 M aqueous solution of HNO, is prepared. Fill in the ICE table with the appropriate value for each involved species to determine concentrations of all reactants and products. HNO,(aq) H,O*(aq) NO, (aq) + (1)O°H Initial (M) Change (M) Equilibrium (M) 5 RESET 0.30 X- 1.4 x 10-11 X+ 7.1 x 10-4 09'0 0.020 - 2r 1.4 x 10-11 + x 1.4 x 10-11 -r 0.60 + 2r -2r X- 09'0 +2x X+ 09°0 1.4 x 10-11 - 2r 1.4 x 10-11 + 2r ) %23 $ 9.How would the addition of a few drops of a NaOH solution affect the equilibrium in system (B)? HC2H3O2(ag)=H+(ag) + C2H3O2- (ag) O No effect will be observed. The hydroxide would neutralize the H+ ions present and shift the equilibrium to the left. O The hydroxide would neutralize the H+ ions present and shift the equilibrium to the right.
- You are given the following information. acid Ка HF 7.2X10-4 HC2H302 1.8×10-5 HCIO 3.5X10-8 HCN 4.0X10-10 Determine the equilibrium constant for each of the following reactions and indicate which arrow would be more appropriate. (Hint: Reactions that produce less than 0.1% of products and reactions that retain less than 0.1% of reactants are not considered to be "equilibrium reactions"). (a) HC2H302 + F C2H302 K = 0 0.025 + HF (b) HCN + CIO - CN + HCIO K = 40 0.010 (c) HCIO + F - CIO + HF K = 40 87.5 lo+ (d) HF + CN + HCN K = 00 1.14e-2Phosphoric acid dissociates to form phosphate anion and hydronium cation. The K, for this reaction is 4.5 x 10-13. Choose the corre equation which expresses this reaction. 4.5 x 10 13 = ([PO4°] x [H3O*j³ ) / ( [H3PO4] × [H2O] ) 4.5 x 10 13 = ([PO4°] × [H3O*] ) / [H3PO4] 4.5 x 1013 = ( [PO4³] × [H3O*]³ ) / [H3PO4] %3D 4.5 x 10 13 = [H3PO4] / ( [PO43-] × [H3O*j³ ) %3D 5 Question 141.- Calculate the value of the equilibrium constant for the following reactions: a) NH3 +H3O+NH4+ + H2O b) NO2^- + H3O+ HNO2 + H2O c) NH4^+ + OH- NH3 +H2O d) HNO2 + OH- NO2^- + H2O e) Classify the reaction in a through d. f) In solving problems, for the reactions a through d, should the final concentration for these reactions be found by treating the reactions as equilibrium or as stoichiometry problems?
- 6. What are the equilibrium concentrations of a 0.80 M HCN solution for each species (in M to two decimal places)? HCN(aq) = H(aq) + CN¯(aq) Kc = 4.9 × 10-10 A: [HCN] = 0.80, [H] = 1.98×105, [CN] = 1.98×10 ghansWhen there is a lot of reactants than of Products at equil, K is ___ = 1 > 1What will happen to [Pb2+(aq)] in the following equilibrium, if NaCl(s) is added to the container? PbCl2(s) < à Pb2+(aq + 2Cl-(aq)