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![+ H₂O - to OH I [ C₂ H₂O₂4] [H₂O*] ГС, наа, с 1 Ka = xxx c ka = x2 [Hz of} = x = veka *C=0.40m,ka=1.4x10² [taot] = 50.40X14X0](http://img.homeworklib.com/questions/2fab6de0-7149-11ea-8f2c-65bc1751ca22.png?x-oss-process=image/resize,w_560)
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3) The pKa of chloroacetic acid is 2.9; the Ka is 1.4 x 103. Complete the...
The pKa of chloroacetic acid is 2.9. If an organic chemist has a 0.10 M aqueous...
The pKa of chloroacetic acid is 2.9. If an organic chemist has a 0.10 M aqueous solution of chloroacetic acid at pH 2.4, what percentage of molecules/ions will be in the basic form?
Equations pH=-log[H3O+]; pOH= -log[OH]; pKw=14.00=pH+POH; Ka-[H3O+][A-[HA]; Kb=[BH+][OH-)[B); pKa=-log Ka; Ka. Kb=Kw Constants Ka (HS-)=1x10-19; Ka (HF)...
Equations pH=-log[H3O+]; pOH= -log[OH]; pKw=14.00=pH+POH; Ka-[H3O+][A-[HA]; Kb=[BH+][OH-)[B); pKa=-log Ka; Ka. Kb=Kw Constants Ka (HS-)=1x10-19; Ka (HF) 7.2x10-4; Ka ([Al(H20).]+)7.9x10-6; Ka (H3PO4)=7.5x10-2: Ka (HPO42-) =3.6x10-13; Ka (HCI)=Huge; Ka(Na+)=tiny; Ka(Cl-)tiny; Kb(NH3)=1.8x10-5; Kw=1x10-14 1) Calculate the pH of aqueous 0.25M HCl and 0.25M HF solutions. 2) Complete the following tables for aqueous solutions of conjugate acid-base pKa pKb Kb 1.3x10-4 35 3) Complete the following table for aqueous solutions DHL TH+) OH) pOH 1x102 4) Calculate the pH of aqueous 0.15M Ba(OH)2 and...
Formic acid Ka: 1.80x10^-4 Propanoic acid Ka: 1.34x10^-5 Chloroacetic acid Ka: 1.36x10^-3 Iodic acid Ka: 1.7x10^-1...
Formic acid Ka: 1.80x10^-4
Propanoic acid Ka: 1.34x10^-5
Chloroacetic acid Ka: 1.36x10^-3
Iodic acid Ka: 1.7x10^-1
What is the pH of a solution that is (This problem requires values in your textbook's specific appendices, which you can access through the OWLv2 MindTap Reader. You should not use the OWLv2 References' Tables to answer this question as the values will not match.) a prepared by dissolving 8.20 g of formic acid (46.03 g/mol) and 10.74 g of sodium formate (68.01 g/mol)...
At the equivalence point of a titration of HOAc (acetic acid, pKa = 1.76 x 10-5)...
At the equivalence point of a titration of HOAc (acetic acid, pKa = 1.76 x 10-5) with NaOH, the species present of OAc" and H20. If the concentration of OAc at the equivalence point is 0.50 M, what is the pH of the solution? Remember that Kb x Ka = 10-14 and that the OAc will react with water as follows: OAC- + H20 --> HOÀc + OH- . 1 4.77 2. 10.23 3 5.68 4. 9.23
11a) 1.00 moles of lactic acid (HC3H5O3, Ka = 1.4 x 10-4) and 1.00 moles of...
11a) 1.00 moles of lactic acid (HC3H5O3, Ka = 1.4 x 10-4) and 1.00 moles of sodium lactate (NaC3H5O3) are dissolved in water resulting in a solution with a final volume of 550 mL. What is the pH of the solution after the addition of 0.080 moles of HCI? a. 0.80 b. 1.8 C. 2.8 d. 3.8
4.13 Just as pH is the negative logarithm of [H3O+], pKa is the negative logarithm of...
4.13 Just as pH is the negative logarithm of [H3O+], pKa is the negative logarithm of Ka, pKa=−logKa The Henderson-Hasselbalch equation is used to calculate the pH of buffer solutions: pH=pKa+log[base][acid] Notice that the pH of a buffer has a value close to the pKa of the acid, differing only by the logarithm of the concentration ratio [base]/[acid]. Part B How many grams of dry NH4Cl need to be added to 2.10 L of a 0.400 M solution of ammonia,...
Consider a titration of 25.00 mL Chloroacetic Acid solution [ka=1.4x10^-3] with 0.1202 M solution of sodium...
Consider a titration of 25.00 mL Chloroacetic Acid solution [ka=1.4x10^-3] with 0.1202 M solution of sodium hydroxide. The volume of 27.40 mL of NaOH(aq) was needed to reach the equivalence point. Calculate: a) The concentration of the chloroacetic acid solution before the titration b) the pH of the chloroacetic acid solution before titration c) the pH of the solution at half equivalence point d) the pH of the solution at the equivalence point e) the pH of the solution when...
The following pictures represent solutions that contain a weak acid HA (pka = 5.0) and its...
The following pictures represent solutions that contain a weak acid HA (pka = 5.0) and its potassium salt KA. Unshaded spheres represent H atoms and shaded spheres represent A-ions. (K+, H30, OH-, and solvent H2O molecules have been omitted for clarity.) -HA -A IC (1) 12) Which of these solutions are buffers? A) (1) and (2) C) (1), (2) and (3) B) (1) and (3) D) All are buffer solutions. following pictures represent solutions at various stages in the titration...
Need help on questions 1-3 Henderson-Hasselbalch: pH=pka + Log( [base]/(acid] ) 1. The value of Ka...
Need help on questions 1-3
Henderson-Hasselbalch: pH=pka + Log( [base]/(acid] ) 1. The value of Ka of nitrous acid (HNO2) is 4.6 x 104 M. Calculate the pH and the concentration of [H3O+] in a 0.02M aqueous solution of HNO2(aq). 2. Label conjugate acid-base pairs: HNO2(aq) + H2O → H30* + NO2 (aq) 3. What will happen to the reaction equilibrium if we increase the pressure in the reaction vessel? H2(g) + 12(e) → 2 HI(g)
i just need 103 & 105 please explain fully 15.103. Predict which solution in each pair...
i just need 103 & 105 please explain fully
15.103. Predict which solution in each pair below will have the a. 1.25 X 10" M HNO3 or 1.00 x 10" M NaOH higher pH. b. 0.345 mM HBrO (PK. = 8.64) or 0.345 mM HCIO 45 mM Be(OH)2 (K = 5 x 10-1) or 45 mM Ba(OH)2 (K. = 2.9 X 10-8) c. d. 1.6 mm (pKs = 8.03) or 0.60 mM (pK6 = 8.32) e. 1.67 X 10-3 M...
Equations pH=-log[H3O+]; pOH= -log[OH]; pKw=14.00=pH+POH; Ka-[H3O+][A-[HA]; Kb=[BH+][OH-)[B); pKa=-log Ka; Ka. Kb=Kw Constants Ka (HS-)=1x10-19; Ka (HF) 7.2x10-4; Ka ([Al(H20).]+)7.9x10-6; Ka (H3PO4)=7.5x10-2: Ka (HPO42-) =3.6x10-13; Ka (HCI)=Huge; Ka(Na+)=tiny; Ka(Cl-)tiny; Kb(NH3)=1.8x10-5; Kw=1x10-14 1) Calculate the pH of aqueous 0.25M HCl and 0.25M HF solutions. 2) Complete the following tables for aqueous solutions of conjugate acid-base pKa pKb Kb 1.3x10-4 35 3) Complete the following table for aqueous solutions DHL TH+) OH) pOH 1x102 4) Calculate the pH of aqueous 0.15M Ba(OH)2 and...
Formic acid Ka: 1.80x10^-4
Propanoic acid Ka: 1.34x10^-5
Chloroacetic acid Ka: 1.36x10^-3
Iodic acid Ka: 1.7x10^-1
What is the pH of a solution that is (This problem requires values in your textbook's specific appendices, which you can access through the OWLv2 MindTap Reader. You should not use the OWLv2 References' Tables to answer this question as the values will not match.) a prepared by dissolving 8.20 g of formic acid (46.03 g/mol) and 10.74 g of sodium formate (68.01 g/mol)...
At the equivalence point of a titration of HOAc (acetic acid, pKa = 1.76 x 10-5) with NaOH, the species present of OAc" and H20. If the concentration of OAc at the equivalence point is 0.50 M, what is the pH of the solution? Remember that Kb x Ka = 10-14 and that the OAc will react with water as follows: OAC- + H20 --> HOÀc + OH- . 1 4.77 2. 10.23 3 5.68 4. 9.23
11a) 1.00 moles of lactic acid (HC3H5O3, Ka = 1.4 x 10-4) and 1.00 moles of sodium lactate (NaC3H5O3) are dissolved in water resulting in a solution with a final volume of 550 mL. What is the pH of the solution after the addition of 0.080 moles of HCI? a. 0.80 b. 1.8 C. 2.8 d. 3.8
The following pictures represent solutions that contain a weak acid HA (pka = 5.0) and its potassium salt KA. Unshaded spheres represent H atoms and shaded spheres represent A-ions. (K+, H30, OH-, and solvent H2O molecules have been omitted for clarity.) -HA -A IC (1) 12) Which of these solutions are buffers? A) (1) and (2) C) (1), (2) and (3) B) (1) and (3) D) All are buffer solutions. following pictures represent solutions at various stages in the titration...
Need help on questions 1-3
Henderson-Hasselbalch: pH=pka + Log( [base]/(acid] ) 1. The value of Ka of nitrous acid (HNO2) is 4.6 x 104 M. Calculate the pH and the concentration of [H3O+] in a 0.02M aqueous solution of HNO2(aq). 2. Label conjugate acid-base pairs: HNO2(aq) + H2O → H30* + NO2 (aq) 3. What will happen to the reaction equilibrium if we increase the pressure in the reaction vessel? H2(g) + 12(e) → 2 HI(g)
i just need 103 & 105 please explain fully
15.103. Predict which solution in each pair below will have the a. 1.25 X 10" M HNO3 or 1.00 x 10" M NaOH higher pH. b. 0.345 mM HBrO (PK. = 8.64) or 0.345 mM HCIO 45 mM Be(OH)2 (K = 5 x 10-1) or 45 mM Ba(OH)2 (K. = 2.9 X 10-8) c. d. 1.6 mm (pKs = 8.03) or 0.60 mM (pK6 = 8.32) e. 1.67 X 10-3 M...
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