Find the pH of the solution containing 0.04 M H3PO4 without neglecting it. Ka1 = 7x10-3, Ka2 = 2.2x10-7, Ka3 = 4.5x10-13) Note: Write the relevant equilibrium reaction, specify the equilibrium constant
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Find the pH of the solution containing 0.04 M H3PO4 without neglecting it. Ka1 = 7x10-3, Ka2 = 2.2x10-7, Ka3 = 4.5x10-13) Note: Write the relevant equilibrium reaction, specify the equilibrium constant.
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- 2&3. Write the expression for Kc (equilibrium constant) for the following reactions.The equilibrium expression for any weak acid can be written as HA (aq) + H20 (1) = A- (aq) + H;O+ (aq) 1. Write the K value expression based on the equation above (remember that pure liquids are not included in the K expression). This is given the special symbol Ka. 2. In this experiment, you will be using pH to find [H3O+]. The relationship is [H3O*] = 10-PH . For a pH of 7.4, find the [H3O+].Write the concentration equilibrium constant expression for this reaction. CH3CO2C2H5(aq)+H2O(l)→CH¸CO₂H(aq)+C2H5OH(aq) ☐
- At equilibrium, an aqueous solution of formic acid (HCHO₂) has these concentrations: [H+] = 0.0029 M; |CHO;] = 0.0029 M; and |HCHO,] = 0.0467 M. HCHO₂(aq) = H+ (aq) + CHO₂ (aq) What is the value of K of this equilibrium? K= Does the equilibrium lie to the right or to the left? O left rightWrite the equilibrium constant expression, K, for the following reaction taking place in dilute aqueous solution. HF (aq) + OH- (aq)F- (aq) + H2O (l) K =Given the following reactions and equilibrium constants: (1) HCN(g) + OH* — CN- + H₂O (g) K₁ =4.9 x 104 (2) H₂O(g) H¹(g) + OH- K₂ = 1.0 x 10-14 - What will be the equilibrium constant for: HCN(g) → CN- + H* (g) ? um constant 5.9 x 104 4.9 x 10-10 5.0 x 10-10 4.9 x 10-40 4.9 x 1018
- For the equilibrium situation involving acetic acid, CH3COOH(aq) + H2O() = CH3CO0 (aq) + H3O*(aq) explain the equilibrium shift that occurs when more acetic acid is added to the solution. Select all that apply. The amount of CH3COO¯ increases. The amount of CH3COO¯ decreases. The amount of H3O* decreases. The equilibrium will shift to the left. The amount of H3O* increases. The equilibrium will shift to the right.When only one concentration at equilibrium is known the concentrations of the others can be calculated using stoichiometry. This is possible because all relationships in a chemical reaction are direct proportions. Calculate the the missing concentrations, then calculate the K, when a 0.050 M solution of CH3NH₂ is made, and at equilibrium [CH3NH₂] = 0.040 M. CH3NH2 + H₂O → CH³NH₂+¹ + OH-¹ 0 0 start: 0.050 M equil: 0.045 M 0.005 M The amount of CH3NH₂ that was converted into product was 0.050 M -0.045 M = For each 1 mol CH3NH3+¹ (0.005 M CH3NH₂ X- Kea 1 mol CH3NH₂ (0.005 M CH3NH2)(- 1 mol OH-1 1 mol CH3NH₂ [CH3NH3 +¹][OH-¹] [ of CH3NH₂ converted into products, there will be 0.005 M of CH3NH3*¹ and -) = ][ CH3NH3+1 OH-1 [CH3NH₂] [ ] (notice that the information to calculate both the pOH and pk, are provided) A. HF B. H₂O C. F-1 D. H30+1 J. 0.005 M K. 0.040 M L. 0.045 M E. CH3NH₂ F. CH3NH₂+¹ M. 0.00056 M N. 0.0056 M G. OH-1 H. 0.030 M I. 0.050 M O. 0.11 M P. 0.50 M of OH-1, orBe sure to answer all parts. Consider the formation of ammonia in two experiments. (a) To a 1.00-L container at 727°C, 1.30 mol of N₂ and 1.65 mol of H₂ are added. At equilibrium, 0.100 mol of NH3 is present. Calculate the equilibrium concentrations of N₂ and H₂ and find Ke for the reaction: [N₂ leq = [H₂ leq 2 NH3(g) = N₂(g) + 3 H₂(g) Kc = ㅈ = M Kc = (b) In a different 1.00-L container at the same temperature, equilibrium is established with -2 8.34 × 10² mol of NH3, 1.50 mol of N2, and 1.25 mol of H₂ present. Calculate Ke for the reaction: M 3 NH3(g) = N₂(g) + H₂(g) 2 2 C C < Prev *********** 18 of 20 MacBook Pro Search or type URL Next *********** ☆
- Write the equilibrium constant expression, K, for the following reaction taking place in dilute aqueous solution. ? NH3 (aq) + H20(1) NH4* (аq) + OH (аq) K =Calculate a value for the equilibrium constant for the reaction (6)*O = (6)0 + (6) ²0 given hv NO2 (9) NO(g) + O(g) K = 2.3 x 10-49 O3 (9) + NO(9) = NO,(g) + O2(9) K = 8.4 x 10-34 (Hint: When reactions are added together, the equilibrium expressions are multiplied.) K =Write down the equilibrium constant expression for the following equation: HF (aq) + H2O (I) = F- (aq) + H3O+ (aq) 2H2 (g) + O2 (g) = 2H2O (g) 2N2O (g) = 2N2 (g) + O2 (g)