Use the References to access important values if needed for this question. The pOH of an aqueous solution of 0.580 M nitrous acid, ( Ka (HNO2) = 4.5 × 10-4) is
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- In the following acid-base equilibria of weak acids in water, label the acid (A), the base (B), the conjugate acid (CA), and the conjugate base (CB). HCIO, (aq) + H,O(1) = H,O*(aq) + CI0, (aq) H,CO, (aq) + H,O1) – H;O*(aq) + HCO; (aq) Answer Bank H,O(1) + CH;NH†(aq) = CH,NH,(aq) + H;O*(aq) СА А B СВ CH, COOH(aq) + H,O(1) - CH;COO (aq) + H;O*(aq)Propionic acid, HC3H5O2, has Ka= 1.34 x 10–5. (a) What is the molar concentration of H3O+ in 0.15 M HC3H5O2 and the pH of the solution? (b) What is the Kb value for the propionate ion, C3H5O2–? (c) Calculate the pH of 0.15 M solution of sodium propionate, NaC3H5O2. (d) Calculate the pH of solution that contains 0.12 M HC3H5O2 and 0.25 M NaC3H5O2.Use the Referenc The pOH of an aqueous solution of 0.313 M acetylsalicylic acid (aspirin), HC9H¬O4, is
- At 50 °C, the ion-product constant for H2O has the valueKw = 5.48 x 10-14. (a) What is the pH of pure water at50 °C? (b) Based on the change in Kw with temperature,predict whether ΔH is positive, negative, or zero for theautoionization reaction of water:2 H2O(l) ⇌ H3O+(aq) + OH-(aq)The base protonation constant K, of allantoin (C4H,N,O,NH,) is 9.12 x 10 °. 9. 3 Calculate the pH of a 0.47 M solution of allantoin at 25 °C. Round your answer to 1 decimal place. pH = | ?What is the pOH for a solution at 25 °C that has a H3O+ concentration of 6.57 ×10-6 M? A 34.1 % (NH4 )2SO4 (molar mass = 132.1 g mol−1) has a density of 1.15 g mL−1. What is the molarity of this solution? (the answer should be entered with 3 significant figures; do not enter units; give answer in normal notation--examples include 1.23 and 120. and -123 and 123. and 12.3) Determine the boiling point of a solution that contains 78.5 g of compound W (molar mass = 132.5 g mol–1) dissolved in 1088.6 g benzene (C6H6; molar mass = 84.156 g mol–1; Kb = 2.53 °C m–1; boiling point of pure benzene = 80.1 °C). (the answer should be entered with 3 significant figures; do not enter units; give answer in normal notation--examples include 1.23 and 120. and -123 and 123. and 12.3) NEED HELP ASAP NO WORK NEEDED
- Please answer these questions on your Page 8 (a) HA(aq) is a weak acid with a dissociation constant, Ka, of 7.7 x 10-¹2. What is the pH of a 0.011 M solution of A-(aq)? The temperature is 25°C. (b) For the reaction A(g) A(1), the equilibrium constant is 0.666 at 25.0°C and 0.111 at 75.0°C. Making the approximation that the entropy and enthalpy changes of this reaction do not change with temperature, at what temperature will the equilibrium constant be equal to 0.777?Use the References to access important values if needed for this question. The pH of an aqueous solution of 0.580 M acetylsalicylic acid (aspirin), (Ka (HC9H7O4) = 3.00 x 10^-4) is(i) Define pH in words. The strong acid HClaq has a pH value of 1, use the following equation for a strong acid: HClaq à H+aq + Cl-aq and convert the following expression to deduce the hydrogen ion concentration: pH = -log10 [H+] (ii) Use the above expression to deduce the pH of HCl (aq) given the concentration of the acid to be 4.5 mol/dm3
- (i) Define pH in words. The strong acid HClaq has a pH value of 1, use the following equation for a strong acid: HClaq à H+aq + Cl-aq and convert the following expression to deduce the hydrogen ion concentration: pH = -log10 [H+] (ii) Use the above expression to deduce the pH of HCl (aq) given the concentration of the acid to be 4.5 mol/dm3 pH =4. How does the pH of each of the following solutions change when 5.0 mL of 1.0 M NaOH (a strong base) is added? Fill in the table. Give your answers with 2 decimals. Initial pH Final pH after adding NaOH Solution (a) 100.0 ml water (b) (c) 100.0 mL 0.150 M HNO2 (a weak acid) (Given: Ka = 4.5 × 10-4) 100.0 mL solution of 0.150 M HNO2 and 0.100 M NaNO₂Given 0.01 M solutions of each of the following acids, which solution would have the lowest pH? -117 Hypoiodous acid (HOI), K = 2.3 x 10 Hypobromous acid (HOBr), K = 2.5 x 10 Lactic acid (HC₂H₂O₂), K = 1.3 x 10 Chlorous acid (HClO₂), K = 1.1 x 10²