What is the concentration of formate ion (CHO2-) in solution at equilibrium, if a 0.250M formic acid (HCHO2) aqueous solution was prepared. The percent ionization for formic acid is 6.22%. Answer:
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We knew that if
HA --->H+ +A-
Percentage of ionization = {[H+]/[HA]}×100
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- A solution contains 2.2 × 10-3 M in Cu2+ and 0.33 M in LiCN. If the Kf for Cu(CN)42- is 1.0 × 1025, how much copper ion remains at equilibrium? Answer: 1.9*10^-26 How do you get that answer can someone show me the steps for solving it?What is the equilibrium concentration of H2CO3 if the initial concentration ofCO2 is 0.546 M and the Kb = 1.3 x 10-3 for the followingequation? Show all work. CO2(aq) + H2O (l) → H2CO3 (aq)A buffer containing acetic acid and sodium acetate has a pH of 5.55. The Ka value for CH3CO2H is 1.80×10^-5. What is the ratio of the concentration of CH3CO2H to CH3CO2^- ?
- In solutions 1-4 you are adding successively larger volumes of 0.00200 M SCN- to the Fe3+ solution and diluting to 10.00 mL. Calculate the final diluted molarity of SCN- in solution #1. 3 sig figs (Calculate the same thing for the solutions 2-4 and enter the results in table 3 as [SCN-] pre-equilibrium.)The Ksp value for magnesium arsenate [Mg3(AsO4)2] is 2.00 X 10-20 so if a chemist added 1.19 x 10-2 M of Pb3(AsO4)2(aq) which is a common ion then what would be the concentration of the Arsenate ion AsO43-(aq) in grams per Liter at equilibrium? Question 15 options: 2.44 3.31 4.47 4.23 2.61 3.02 3.78 3.96 2.87 3.57The Ka for benzoic acid is 6.5 x 10-5. What is the pH of a 0.030 M benzoic acid solution? (note: please express your answer with 3 sig figs.; the equilibrium reaction is shown below:)
- Please answer Determine the volume of the stock solutions of 1.00 mol L-1 acetic acid and 1.00-mol L-1 of sodium acetate needed to prepare the solutions. Write your answer in the table below. Calculate for the initial pH of the solutions and write the pH in the table below. Which solution(s) have the same pH? Which solution is closest to the pKa? Calculate the pH of the solutions and the change in pH after adding 5.00 mL of 0.10- mol L-1 NaOH. Calculate the pH of the solutions and the change in pH after adding 5.00 mL of 0.10- mol L-1 HCl. Which solution is most resistant to pH change? Which solution is least resistant to pH change?ANSWER D,E & FIn some natural systems the pH must be maintained within very narrow limits. e.g. in human blood the pH must remain close to 7.4 or cell deterioration occurs. Blood contains several weak acid/conjugate base equilibria called buffers, which control the pH.One weak acid present in blood is the dihydrogen phosphate ion, H2PO4-(aq), for which the equilibrium in aqueous solution is H2PO4-(aq) +H2O(l) <--> HPO42-(aq) + H3O+(aq). a) What would be the effect on this equillibrium of adding hydrochloric acid solution?b) Write the expression for the equilibrium constant of this reaction.c) If 0.5 mol H2PO4- and 0.5 mol HPO42- are in equilibrium in 1.0L of aqueous solution, calculate the pH of the solution. (Ka for H2PO4-= 6.4 x 10-8)d) 0.010 mol HCl is now added to the 1.0L of the solution in c). Assuming that all the added H+ ions are used up in the equilibrium shift, calculate the concentrations of H2PO4- and HPO42-. Hence calcuate the pH of this solution.e) 0.010 mol…Will a precipitate form when 100.0 mL of a 4.6 ✕ 10-4 M Mg(NO3)2 is added to 100.0 mL of 5.1 ✕ 10-4 M NaOH? (Hint: Ksp for Mg(OH)2 = 8.9 ✕ 10-12) Yes Give the value of Q, to support your answer. __________________________________
- You take 10.00 grams of pure acetic acid add top it up to 1.00 liter of solution. What is the pH of the solution at equilibrium? You will need to look up the pKa or Ka of acetic acid to figure this out (just google it). You add 20.00 grams of sodium acetate to the mixture and it causes a negligible increase in volume. What is the new pH once the solution reaches equilibrium? You add 1.00 mole of HClO4 (a strong acid) causing a negligible increase in volume again. What is the new pH? You add 1.00 mole of NaOH (strong base) causing a negligible increase in volume AGAIN (somehow?). What is the new pH?Question 3Reaction 1:PbCl2(s) ↔Pb2+(aq) + 2Cl−(aq) Ksp=K1 Reaction 2:AgCl(s) ↔ Ag+(aq) + Cl−(aq) Ksp=K2Based on the information given above, which of the following is the expression for Keq (K3)for the reaction that occurs when a0.1MAgNO3(aq) is added to a saturated solution of PbCl2(aq), as represented by the following chemical equation?PbCl2(s)+2Ag+(aq)⇄Pb2+(aq)+2AgCl(s) a) Keq=K3=K1 + (2 x K2)b) Keq=K3= K1 - (2 x K2)c) Keq=K3= K1 x (K2)2d) Keq=K3= K1/(K2)24. In question 3 if Ksp for reaction 1 is 1.6 x 105 and the Ksp for reaction 2 is 1.6 x 10−1o, then what is the value for K3 or Keq? Do number 4If 21 mL of 3.2 x 10–5 M magnesium chloride and 15 mL of 1.5 x 10–4 M sodium fluoride are mixed, what are the concentrations of magnesium chloride and sodium fluoride after mixing? Keep an extra sig fig. Question 8 options: sodium fluoride magnesium chloride 1. 1.87 x 10–5 M 2. 3.73 x 10–5 M 3. 6.25 x 10–5 M 4. 1.25 x 10–4 M 5. 1.50 x 10–4 M