101 Che XC Det XC Det XWha XAns XC Wh XC Anl XX Mat X18.1 Xb Ans Xb Ans XSea X+app.101edu.coS O2dUC mailCanvasI UC microsoft108" Wide Quilt Bac...14 Google Calendar-.101 Chem101b My Solutions | bartl..Google DocsQuestion 27 of 52SubmitDetermine the pH of a 0.30 M solution of HF (Ka = 6.8 × 10-4).%3D( PREV123NEXT >Based on your ICE table and definition of Ka, set up the expression for Ka in order todetermine the unknown. Do not combine or simplify terms.Ka =6.8х 10-42 RESET[0][0.30][6.8 x 10-4][0.15][x][2r][2r]?[0.30 + x][0.30 - x][0.30 + 2x][0.30 - 21][6.8 x 10-4 + x][6.8 x 10-4 - x][6.8 x 10-4 + 2x][6.8 x 10-4 - 2r]+M 9US17:30... 101 Che XC Det XC Det XWha XAns XC Wh XC Anl XX Mat X18.1 Xb Ans Xb Ans XSea X+app.101edu.coS O2dUC mailCanvasI UC microsoft108" Wide Quilt Bac...14 Google Calendar-.101 Chem101b My Solutions | bartl..Google DocsQuestion 27 of 52SubmitDetermine the pH of a 0.30 M solution of HF (Ka = 6.8 × 10-4).%3D13NEXT >Based on the given values, set up ICE table in order to determine the unknown.HF(aq)H,O(1)H,O*(aq)F(aq)++Initial (M)Change (M)Equilibrium (M)2 RESET0.306.8 x 10-40.15+x-X+2r-2x0.30 +x0.30 - x0.30 + 2x0.30 - 2x6.8 × 10-4 + x6.8 × 10-4 - x6.8 × 10-4 + 2r6.8 × 10-4 - 2x+M 9US17:30...

Concentration of HF = 0.30 M Dissociation constant of HF (Ka) = 6.8 x 10-4. The dissociation of HF in water is given as: HFaq + H2Ol ↔ H3O+aq + F-aq In a solution form water is present in excess amount. So we neglect the concentration of water . ICE table is: HF(aq) H2O(aq) H3O+(aq) F-(aq) Initial 0.30 - 0 0 Change -x - +x +x Equilibrium 0.30-x - +x +x Ka=H3O+aq F-aqHFaqSubstituting the values at equilibrium of ICE table in the above equation Ka=[x] [x][0.30-x]=6.8×10-4 Therefore, the expression for Ka is as above.

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Determine the pH of a 0.30 M solution of HF (Ka = 6.8 × 10-4). < PREV 2 3 NEXT > 1 Based on your ICE table and definition of Ka, set up the expression for Ka in order to determine the unknown. Do not combine or simplify terms. Ka = 6.8 × 10-4 5 RESET [0] [0.30] [6.8 x 10-9 [0.15) [24] [0.30 + x] [0.30 - x] [0.30 + 2x] [0.30 - 21] (6.8 × 104 + x] [6.8 x 104 - x) [6.8 × 104 + 2x] [6.8 × 104 - 21]

Determine the pH of a 0.30 M solution of HF (Ka = 6.8 × 10-4). < PREV 2 3 NEXT > 1 Based on your ICE table and definition of Ka, set up the expression for Ka in order to determine the unknown. Do not combine or simplify terms. Ka = 6.8 × 10-4 5 RESET [0] [0.30] [6.8 x 10-9 [0.15) [24] [0.30 + x] [0.30 - x] [0.30 + 2x] [0.30 - 21] (6.8 × 104 + x] [6.8 x 104 - x) [6.8 × 104 + 2x] [6.8 × 104 - 21]

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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