Calculate the Ka for a monoprotic acid, HX, if a 0.100000 M solution has a pH of 3.333. The data is provided. If it wasn'tyou could get the concentrations by converting the pH into [H3O+¹] using [H3O+1] = 10-PH = 10-3.333 =+start:equil:HX + H₂O S0.100000 M0.099536 M0.000464 MX-100.000464 MH3O+100.000464 Mmol ofThe stoichiometry tells us that if we made 0.000464 mol of H3O+¹ we also madeX-1The amount of HX at equilibrium is 0.100000 M HX-M HX][[]D([KaKaa. acetich. CaCO3(s)i. OH-¹-10. H₂PO4¹ p. CO3-²v. H₂CO3bb. 0.1000W. PO4-³CC. 3hh. 2.16 x 10-6b. X-1c. HXd. Al+3j. H30+1q. HCO3-1x. HC₂H3O2dd. 2ii. 1.60 x 10-5k. SO4²r. H₂Se. CO₂ee. 1jj. 0.2000y. C₂H30₂-11. Mg+2S. HS-1f. CaF2(aq)ff. 0.099536m. Cl-¹t. S-²z. 1 x 10-14MX-1 =kk. 6.600 x 10-³g. HF(aq)n. HPO4²u. H₂O(liq)aa. 0.000464gg. 3.95 x 10-3

The given reaction is : HX + H2O ⇔ X- + H3O+ The answer is as follows:

Calculate the K₂ for a monoprotic acid, HX, if a 0.100000 M solution has a pH of 3.333. The data is provided. If it wasn't you could get the concentrations by converting the pH into [H3O+¹] using [H3O+1] = 10-PH = 10-3.333 = H3O+1 start: equil: HX + H₂O 0.100000 M 0.099536 M 0.000464 M X-1 + 0 0.000464 M 0 0.000464 M mol of The stoichiometry tells us that if we made 0.000464 mol of H3O+1 we also made X-1 The amount of HX at equilibrium is 0.100000 M HX- M HX |][ [ x [ Ka = Ka = a. acetic h. CaCO3(s) 0. H₂PO4¹ v. H₂CO3 bb. 0.1000 hh. 2.16 x 10-6 b. X-1 i. OH-¹ P. CO3-2 W. PO43 CC. 3 c. HX d. Al+3 j. H30+1 q. HCO3-1 x. HC₂H3O2 dd. 2 ii. 1.60 x 10-5 k. SO4² r. H₂S e. CO₂ f. CaF2(aq) 1. Mg+2 S. HS-1 y. C₂H30₂-1 ee. 1 ff. 0.099536 jj. 0.2000 m. C-1 t. 5-² Z. 1x 10-14 MX1 = kk. 6.600 x 10-3 g. HF(aq) n. HPO4² u. H₂O(liq) aa. 0.000464 gg. 3.95 x 10-3

Calculate the K₂ for a monoprotic acid, HX, if a 0.100000 M solution has a pH of 3.333. The data is provided. If it wasn't you could get the concentrations by converting the pH into [H3O+¹] using [H3O+1] = 10-PH = 10-3.333 = H3O+1 start: equil: HX + H₂O 0.100000 M 0.099536 M 0.000464 M X-1 + 0 0.000464 M 0 0.000464 M mol of The stoichiometry tells us that if we made 0.000464 mol of H3O+1 we also made X-1 The amount of HX at equilibrium is 0.100000 M HX- M HX |][ [ x [ Ka = Ka = a. acetic h. CaCO3(s) 0. H₂PO4¹ v. H₂CO3 bb. 0.1000 hh. 2.16 x 10-6 b. X-1 i. OH-¹ P. CO3-2 W. PO43 CC. 3 c. HX d. Al+3 j. H30+1 q. HCO3-1 x. HC₂H3O2 dd. 2 ii. 1.60 x 10-5 k. SO4² r. H₂S e. CO₂ f. CaF2(aq) 1. Mg+2 S. HS-1 y. C₂H30₂-1 ee. 1 ff. 0.099536 jj. 0.2000 m. C-1 t. 5-² Z. 1x 10-14 MX1 = kk. 6.600 x 10-3 g. HF(aq) n. HPO4² u. H₂O(liq) aa. 0.000464 gg. 3.95 x 10-3

Chemistry: Principles and Practice

3rd Edition

ISBN:9780534420123

Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer

Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer

Chapter16: Reactions Between Acids And Bases

Section: Chapter Questions

Problem 16.30QE

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Similar questions

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