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13) (5 pts) Calculate the pH (aq., 25 °C) of a 0.0150 M chloroacetic acid (CH,CICOOH)...
Calculate the Ka (aq., 25 °C) of a 0.250 M weak acid whose pH is 3.86. The base-ionization constant of ethylamine (C2H5NH2) is 6.4 x 10-4 at 25.0 °C. The [H+) in a 1.6 x 10-2 M solution of ethylamine is M.
5.) Calculate the concentrations of all the species and the pH in 0.25 M hypochlorous acid, HOCL. For HOCL, Ka=3.5x10^-8 6.) the pH of a 0.115M solution of chloroacetic acid, CICH2COOH, is measured to be 1.85. Calculate the Ka for this monoprotic acid Calculate the concentrations of all the species and the pH in 0.25 M hypochlorous acid, HOCI. For HOC, Ka :3.5 x 108. .) The pH of a 0.115M solution of chloroacetic acid, CICH2COOH, is measured to be...
Calculate the pH of a 7.74×10-5 M solution of hypochlorous acid at 25°C. Ka for hypochlorous acid is 3 x 10−8.
The pH of a 0.115 M solution of the weak acid, chloroacetic acid (chemical formula ClCH2COOH), is measured to be 1.92 at equilibrium. Calculate the Ka of this monoprotic acid.
10 pts Question 2. Suppose 5*10-4 M chloroacetic acid (a) What are the sources of proton? Which proton source can be ignored and why? (b) What is the pH of the solution? (c) Show and confirm any assumptions made in calculation of pH. (d) What is analytical concentration of chloroacetic acid?
An initial solution is made by combing 13 mL of 0.8 M chloroacetic acid with 20 mL of a NaCH2ClCOO solution. (Ka= 1.4x10-3) a) The pH of the resulting solution was measured to be 2.92. What is the concentration of the NaCH2ClCOO solution? b) The solution from part a) is then titrated with 5 mL of 0.8 M NaOH. What is the pH of the resulting solution? c) If the solution is then titrated to its equivalence point, how much...
1. Calculate the pH at 25°C of a 0.757 M solution of a weak acid that has Ka = 9.28 x 10 2. Determine the Kb of a weak base if a 2.50 M solution of the base has a pH of 9.595 at 25°C. 3. The overall dissociation of malonic acid, H2C3H2O4, is represented below. The overall dissociation constant is also indicated. H2C3H204 = 2 H+ + C3H2042- K = 3.0 x 10-9 To a 0.025-molar solution of malonic...
The Ka value for acetic acid, CH3COOH(aq), is 1.8×10−5. Calculate the pH of a 2.40 M acetic acid solution. pH= Calculate the pH of the resulting solution when 2.50 mL of the 2.40 M acetic acid is diluted to make a 250.0 mL solution. pH=
4. Please calculate the resulting pH when 25 mL of 0.10 M acetic acid is titrated with 10 mL of 0.10 M NaOH (K. acetic acid = 1.8 x 10-5): CH,COOH (aq) + NaOH(aq) → CH3COONa (aq) + H20 (1)
A solution is prepared at 25° that is initially 0.39M in chloroacetic acid HCH2ClCO2, a weak acid with =Ka×1.310−3, and 0.064M in potassium chloroacetate KCH2ClCO2. Calculate the pH of the solution. Round your answer to 2 decimal places