Concept explainers
Consider the reaction of sodium oxalate
In an oxidation-reduction titration, the
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a. What is the balanced oxidation half-reaction?
b. What is the balanced reduction half-reaction?
c. What is the balanced ionic equation for the reaction?
d. If
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Basic Chemistry
- 5) What is the pH of a 0.250 M HF solution? The K, of HF is 3.5 X 104. (2.03) HF (aq) + H2O (1) H30* (aq) +F ° (aq)arrow_forward(7.2b) Complete and balance the following by ion-electron method. Reactions are taking place in alkaline solution. b) Cr(OH)a) + CIO (nq) - Cro, (aq) Analysis: Balancing under acidic condition first (adding H30 & H") Oxidation-Half reation: Balancing atoms & charges: Cr(OH)x) * H20 - Cro, ug * H' + Reduction-Half reation: Balancing atoms and charges: 10 taq H20 Balancing electrons lost & gained Multiply oxidation-half reaction by Multiply reduction-half reaction by e's Final balanced equation: Continue next page.arrow_forward(7.2b) Complete and balance the following by ion-electron method. Reactions are taking place in alkaline solution. b) Cr(OH)a) + CIO (aq) - Cro, (aq) + Clug Analysis: Balancing under acidic condition first (adding H30 & H*) Oxidation-Half reation: Balancing atoms & charges: Cr(OH)) * H20 - Cro, ug) + Reduction-Half reation: Balancing atoms and charges: e Clap H20 Balancing electrons lost & galned Multiply oxidation-half reaction by e's Multiply reduction-half reaction by e's Final balanced equation: Continue next page.arrow_forward
- (7.2b) Complete and balance the following by ion-electron method. Reactions are taking place in alkaline solution. b) Cr(OH)a) + Clo (aq) - Cro, 2(aq) • Clzup Analysis: Balancing under acidic condition first (adding Hy0 & H") Oxidation-Half reation: Balancing atoms & charges: Cr(OH)y) * H20 - Cro, ng) + H' Reduction-Half reation: Balancing atoms and charges: CIo (aq) e Clp H20 Balancing electrons lost & gained Multiply oxidation-half reaction by e's Multiply reduction-half reaction by e's Final balanced equation: Continue next page.arrow_forward(7.1a) Complete and balance the following by ion-electron method. Reactions are taking place in acidic solution. a) Cu(s) + NO3 (aq) - Cu*2(aq) + NO2(g) + NO29) Analysis: Oxidation-Half reation: Balancing atoms & charges: Cu(s) Cu*2, (aq) e Reduction-Half reation: Balancing O & H and charges: H* + NO3 (aq) e NO2(g) + 1 H20 Balancing electrons lost & gained and balancing the equation: Cu(s) NO3 (aq) + H* - Cu*(aq) NO2(g) H20arrow_forward8) What is the pH of a solution containing 0.280 M weak base, B', with a K, of 3.9 X 107? (10.51) в (ад) + H-0 (1) ВН (аq) + ОН (ад)arrow_forward
- (5.7: Similar to For practice 5.11)The titration of a 20.0 mL sample of an H₂SO4 solution of unknown concentration requires 18.88 mL of a 0.203 M KOH to reach the equivalence point. What is the concentration (in M) of the unknown H₂SO4 solution? (Hint: Write the balanced reaction equation first.) O 0.0958 M O 0.383 M O 0.767 M O 0.192 Marrow_forwardGiven the following equilibrium constants at 433°C, 1 Na20(s) = 2 Na(1) +02 (9) K1 = 5 x 10-25 2 1 NaO(g) - Na(1)+;02(9) K2 8 x 10-5 -29 Na2 O2 (s) = 2 Na(1) + O2 (9) NaO2 (s) = Na(1) + O2 (9) K3 = 4 × 10 -14 KĄ 3 x 10 determine the values for the equilibrium constants for the following reactions: a. Na20(s) + 1 O2 (g) = Na2 O2 (8) Equilibrium constant = b. NaO(g) + Na, 0(s) = Na2O2(s) + Na(1) Equilibrium constant c. 2 NaO(g) = Na2 O2 (s) Equilibrium constant = (Hint: When reaction equations are added, the equilibrium expressions are multiplied.)arrow_forward7. (8 pts) Calculate the solubility product, Ksp, for SrF2, given that its solubility is 0.0080 g per 225.0 mL of solution. SrF (s) Sr2 (aq) 2F (aq) 2arrow_forward
- Write the net ionic equation for the reaction of potassium acetate and aluminum iodide. Al(aq) + C2H3O2(aq) ⟶ AlC2H3O2(s) Al(aq) + 3C2H3O2(aq) ⟶ Al(C2H3O2)3(s) Al+3(aq) + 3C2H3O2-1(aq) ⟶ Al(C2H3O2)3(s) Al+3(aq) + C2H3O2-1(aq) ⟶ Al(C2H3O2)3(s) Al+3(aq) + C2H3O2-1(aq) ⟶ AlC2H3O2(s)arrow_forward1.4 4 (a) Calculate the value of Kc for the reaction: PC15 (8) PC13 (g) + Cl2 (g) AH = Positive Given that when 8.4 mol of PC15 (g): is mixed with 1.8 mol of PC13 (g) and allowed to come to equilibrium in a 10 dm³ container the amount of PC15 (g) at equilibrium is 7.2 mol. Kc = (b) Explain the effect of the following changes below on the value of Kc: (i) Increasing temperature (ii) Lowering the concentration of chlorine (C1₂) (iii) Addition of a catalystarrow_forward(7 points) Calculate the silver ion concentration, [Ag*], of a solution prepared by dissolving 8.25 g of AgNO3 and 91.5 g of KCN in sufficient water to make 1.00 L of solution. Assume the reaction goes to completion. Ag (aq) +2CN (aq) Ag(CN)2(aq) kr = 1.0 x 1021arrow_forward
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