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[C6H6O62-] = M.
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- 6. (i) Define pH in words. The strong acid HClahas a pH value of 1, use the following equation for a strong acid: H+, + 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 (ag) given the concentration of the acid to be 4.5 mol/dm3 pH =6. (i) Define pH in words. The strong acid HClag has a pH value of 1, use the following equation for a strong acid: HClaa Hog + and convert the following expression to deduce the hydrogen ion concentration: pH = -log10 [H+] Clea (ii) Use the above expression to deduce the pH of HCl (ag) given the concentration of the acid to be 4.5 mol/dm³ pH =3. A 0.0560 g quantity of acetic acid is dissolved in enough water to make 50.0 mL of solution. Calculate the concentrations of H;O*, CH;COO and CH;COOH at equilibrium. What is the pH of the solution? (Ka = 1.8x10) CH;COOH(aq) = CH;COO (aq) + H;O* (aq) a) Calculate the initial concentration of CH;COOH. (C:12; H:1; 0:16) b) Calculate the concentration of CH;CoO (aq) and H;O* (aq) at equilibrium. c) Calculate pH of the solution.
- Calculate the pH and the equilibrium concentrations of HC6H6O6- and C6H6O62- in a 0.0669 M ascorbic acid solution, H2C6H6O6 (aq).For H2C6H6O6, Ka1 = 7.9×10-5 and Ka2 = 1.6×10-12Ph= [hc6h6o6] = M [c6h6o6^2-]= M. Please type answer not write by hend2. A typical cup of tea (150.0 mL) is prepared. One teaspoon of glucose (C6H12O6; m.w. = 180.156 g mol-¹) is added (approximately 4.0 g of sugar). The resulting volume of solution is 152.6 mL, your standard cup of tea. Treating the sugar as the solute and the tea as water, the solvent. Calculate the following concentrations: a) M b) m c) % by mass d) mol fractionCalculate the pH and the equilibrium concentrations of HC6H6O6- and C6H6O62- in a 0.2410 M ascorbic acid solution, H2C6H6O6 (aq).For H2C6H6O6, Ka1 = 7.9×10-5 and Ka2 = 1.6×10-12 pH = [HC6H6O6-] = M [C6H6O62-] =
- 10. If 29.50 mL of 0.175 M nitrous acid neutralizes 50.0 ml. of harium hydroxide, what is the molarity of the base? 2HNO;(aq) + Ba(OH);(aq) Ba(NO:)(aq) + 2H;O(1)(b) Sodium hydroxide reacts with propanoic acid in the following equation: NaOH + CH3CH,COOHCH3CH2COONA + HLO A buffer solution is formed when sodium hydroxide is added to an excess of aqueous propanoic acid. (i) Calculate the number of moles of propanoic acid in 50.0 cm of 0.125 mol dm aqueous propanoic acid. (ii) Use your answers to part (b)(i) to calculate the number of moles of propanoic acid in the buffer solution when 2.00 cm' of 0.500 mol dm aqueous sodium -3 hydroxide are added to 50.0 cm' of 0.125 mol dm aqueous propanoic acid. ) Hence calculate the pH of this buffer solution at 298.15 K.An acid-base equilibrium system is created by dissolving 0.50 mol CH3CO2H in water to a volume of 1.0 L. What is the effect of adding 0.50 mol CH3CO2–(aq) to this solution? 1.The pH of the solution will equal 7.00 because equal concentrations of a weak acid and its conjugate base are present. 2.Some CH3CO2H(aq) will ionize, increasing the concentration of CH3CO2–(aq) and increasing the pH.3.Some CH3CO2–(aq) will react with H3O+, increasing the concentration of CH3CO2H(aq) and reestablishing the solution equilibrium. b. 2 only c. 3 only d. 1 and 3 e. 1, 2, and 3
- 1) Calculate the pH (aq., 25 °C) of a 0.073 M Novocaine (C13H21O2N2CI) solution, which is the salt of the base procaine and hydrochloric acid. The pKb of procaine = 5.155. oltion (lom 2) Calculate the pH (aq., 25 °C) of a 0.100 M potassium hypochlorite solution. The pKa of hypochlorous acid = 7.54.For the reaction of hydrazine (N,H4) in water, H2NNH2 (aq) + H2O(1) = H2NNH3+ (aq) + OH (ag) K, is 3.0 x 10. Calculate the concentrations of all species and the pH of a 2.9-M solution of hydrazine in water. [H, NNH2 H2NNH3+ =| [OH ]=| =[_HO] [H*] =[ pH =| %3DCalculate the pH at 25 °C of a 0.15M solution of lidocaine HC1 (C4H2,NONH,CI). Note that lidocaine (C14H2,NONH) is a weak base with a p K, of 7.94. Round your answer to 1 decimal place. pH = | olo Ar
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