1. Write the charge balance for a MgSO4 and NaNO3 solution that contains Ht, OH, Mg²+, HSO4, SO²-4, MgHSO*4,MgOH+, Na+, and NO 3.A. [SO2-4]+[Mg2+]=[OH-]+[HSO−4]+[H+]+[NO−3]+[MgHSO+4]+[MgOH+]+[Na+]B. [H+]+2[Mg2+]+[MgHSO+4]+[MgOH+]+[Na+]=[OH-]+[HSO-4]+2[SO2-4]+[NO-3]C. [H+]+[Mg2+]+[MgHSO+4]+[MgOH+]+[Na+]=[OH-]+[HSO-4]+[SO2-4]+[NO-3]D. [Mg2+]+[MgHSO+4]+[MgOH+]=[OH-]+[HSO−4]+[SO2−4]+[NO−3] +[H+]+[Na+]2. Select each side of the mass balance equation for an Fe3(PO4)2 solution where the species in solution are Fe2+, PO3-4,HPO42-, H2PO4-, H3PO4, and FeOH+.A. 3([Fe2+]+[FeOH+])=3[PO3-4]+2[HPO2-4]+[H2PO-4]B. 3([Fe2+]+[FeOH+])=[PO3-4]+[HPO2-4]+[H2PO-4]+[H3PO4]C.[Fe2+]+[FeOH+]=2([PO3—4]+[HPO2-4]+[H2PO-4]+[H3PO4])D. 2([Fe2+]+[FeOH+]) = 3([PO3-4]+[HPO2-4]+[H2PO-4]+[H3PO4])3. Determine the molar solubility (S) of Ag2CO3 in a buffered solution with a pH of 5.541 using the systematic treatment ofequilibrium. Ksp(Ag2CO3)=8.46x10-12; Ka1(H2CO3)=4.45x10-7; Ka2(H2CO3)=4.69x10-11Use the systematic treatment of equilibrium to solve this problemPlease & Thank you in advance!

“Since you have posted multiple questions, we will provide the solution only to the first question…

1. Write the charge balance for a MgSO4 and NaNO3 solution that contains H+, OH, Mg2+, HSO4, SO²-4, MgHSO*4, MgOH+, Na+, and NO 3. A. [SO2-4]+[Mg2+]=[OH-]+[HSO−4]+[H+]+[NO−3]+[MgHSO+4]+[MgOH+]+[Na+] B. [H+]+2[Mg2+]+[MgHSO+4]+[MgOH+]+[Na+]=[OH-]+[HSO-4]+2[SO2-4]+[NO-3] [H+]+[Mg2+]+[MgHSO+4]+[MgOH+]+[Na+]=[OH-]+[HSO-4]+[SO2-4]+[NO-3] C. D. [Mg2+]+[MgHSO+4]+[MgOH+]=[OH-]+[HSO-4]+[SO2−4]+[NO−3] +[H+]+[Na+]

1. Write the charge balance for a MgSO4 and NaNO3 solution that contains H+, OH, Mg2+, HSO4, SO²-4, MgHSO*4, MgOH+, Na+, and NO 3. A. [SO2-4]+[Mg2+]=[OH-]+[HSO−4]+[H+]+[NO−3]+[MgHSO+4]+[MgOH+]+[Na+] B. [H+]+2[Mg2+]+[MgHSO+4]+[MgOH+]+[Na+]=[OH-]+[HSO-4]+2[SO2-4]+[NO-3] [H+]+[Mg2+]+[MgHSO+4]+[MgOH+]+[Na+]=[OH-]+[HSO-4]+[SO2-4]+[NO-3] C. D. [Mg2+]+[MgHSO+4]+[MgOH+]=[OH-]+[HSO-4]+[SO2−4]+[NO−3] +[H+]+[Na+]

Chemistry & Chemical Reactivity

9th Edition

ISBN:9781133949640

Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel

Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel

Chapter16: Principles Of Chemical Reactivity: The Chemistry Of Acids And Bases

Section: Chapter Questions

Problem 108GQ

See similar textbooks

Similar questions

[OH-] = = [+H] M M
19. Write charge balance equation for 0.1 M HF solution. O [H3O+]-[H30+]=[OH-]=[F-] [H30+]=[OH-1+[F-] [H30+]=[F-] [H30+]^2=[OH-]+[F-]
5. A typical aspirin tablet contains 325mg of acetylsalicylic acid (HC9H7O04) which is a weak acid. The Ka of this acid is 3.69 x 10-4. a. Give the equation for dissociation of the weak acid HC9H7O4. b. What is the initial concentration, in units of molarity, of acetylsalicylic acid if 2 tablets are dissolved in 500. mL of water? c. What is the equilibrium concentration of all species of this solution of acetylsalicylic acid? d. What is the pH, pOH, and % ionization of this solution?
Show your work, use general chemistry rules for significant figures. 6. A solution is made by dissolving 1.0 x 10–4 mol of Na3PO4 in 200.00 mL of water. Then 0.25 mL of 0.200 F HCl is added to the solution. H3PO4 ⇌ H+ + H2PO4– Ka1 = 7.11 x 10–3 H2PO4– ⇌ H+ + HPO42– Ka2 = 6.34 x 10–8 HPO42– ⇌ H+ + PO43– Ka3 = 4.17 x 10–13 (a) What is the pH of the resultant solution? (b) An additional 0.75mL of 0.200 F HCl is added to the solution from the end of part (a). What is the pH of the resultant solution? (c) Finally, an additional 0.50 mL of 0.200 F HCl is added to the solution from the end of part (b). What is the pH of the resultant solution?
Calculate the pH of a 0.0488 M aqueous solution of acetylsalicylic acid (aspirin) (HC9H7O4, Ka = 3.0x 10-4). pH = _________ The hydronium ion concentration of an aqueous solution of 0.429 M acetylsalicylic acid (aspirin) HC9H7O4, Ka = 3.0x 10-4) is [H3O+] = _______ M
What are the concentrations of H3O* and OH in each of the following? а. 2.78 M NaOн O[OH ] = 2.78 M, H30* = 2.78 M ООн ]— 2.78 м, Н;0+| — 3.60 x 10-15 М O[OH] = 0.360 M, H;O+ = 2.78 M 0Он ]— 3.60 х 10-15 М, Нз0*| M, H,0* = 2.78 M b. 0.320 M Sr(ОН)2 O[OH ]= 0.320 M, H30+ = 0.320 M O[OH] = 0.640 M, H3O* = 0.320 M oОн ]— 1.56 х 10-14 М, Н,0+| — 0.320 М [OH ] = 0.640 M, H;O* 1.56 х 10-14 М с. 4.50 х 10 2 МHCIО4 O[OH] = 4.50 x 10-² M, H3O+ = 4.50 × 10-2 M ООн ]— 2.22 х 10-13 М, |НзО+| — 4.50 x 10-2 М [1,0*] = ООн ]— 2.22 х 10-13 М, |НзО+| — 2.22 х 10-13 м %3 4.50 х 10-2 М, H;О*| 3D 2.22 х 10-13 М d. 0.850 M HCI он ] — 0.850 М, Нsо3 0.850 м ООн ]3 1.18 М, |Нә0*| 3 0.850м O[OH] = 1.18 x 10-14 M, |H3O+ = 0.850 M oОн ]3 0.850 м, Нз0*| %3D 1.18х 104 м
The value of Ka1 and Ka2 for oxalic acid (H2C2O4) are 5.90×10-2 and 6.40×10-5 , respectively. (Use H3O+ instead of H+.) Write the equation for the reaction that goes with Ka1: + + Write the equation for the reaction that goes with Ka2: + +
Which acid is the strongest? a. propanoic acid, HC3H5O2, Ka = 1.4×10–5 b. oxalic acid, H2C2O4, K a = 5.9 x 10-2 c. formic acid, HCOOH, Ka = 1.8×10–4 d. hydrofluoric acid, HF, Ka = 3.5 x 10-4
Complete the balanced molecular reaction for the following weak base with a strong acid: 2 NaCIO(aq) + H₂SO4 (aq) → 2 HCIO₂ + Na₂SO4 1 1 + Reset 4- 2 OH H 0₂ N U 3- 3 ☐3 ) ~ 4 4 H₂O H3O+ LO 5 05 14 ☐ 6 0 CO 02+ 7 07 CI U (s) (1) Na 3+ □ 8 9 4+ • x H₂O 0 09 口。 (g) (aq) S Delete
A solution is 0.26 M Sr(OH)2. What are the concentrations of H3 OT and OH in this solution? [Он ] 3D 0.52 М, [НзО*]— 1.9х 10-14 м [Он ] — 0.52 х 10-14 М, (Н;О*]-1.9 M %3| [Он ] — 0.26 М, [НзО*] 3D 0.26 М [ОН ] —D 0.26 М, [Н;О*] — 0.52 М %3D
Determine the concentrations of the ionic species present in a 0.430 M solution of the Na2CO3. (pKa1=6.35, pKa2=10.33 for H2CO3). [H2CO3] [HCO3-] [CO32-] [H3O+] [OH-]
Calculate the concentration of free Ca²⁺ in a solution of 0.10 M CaY²⁻ at pH 10.00. (K= 5.0 × 10¹⁰)