9. Acetic acid (CH,CO,H) asap. of 4 M solution of acetic acid is titrated with NAOH...
9. Acetic acid (CH3CO,H) has a pk of 4.76 at 25°C. A 5.00 mL aliquot of a 0.0185 M solution of acetic acid is titrated with a 0.1000 M standardized solution of NaOH. (a) What is the pH of the acetic acid solution before any of the standardized solution is added? (b) What volume of the standardized solution is necessary to reach the equiv- alence point? (c) What is the pH of the solution after 1.48 mL of the standardized...
9. Acetic acid (CH3CO2H) has a pk of 4.76 at 25°C. A 5.00 mL aliquot of a 0.0185 M solution of acetic acid is titrated with a 0.1000 M standardized solution of NaOH. (a) What is the pH of the acetic acid solution before any of the standardized solution is added? (b) What volume of the standardized solution is necessary to reach the equiv- alence point? (c) What is the pH of the solution after 1.48 mL of the standardized...
A 35.0-mL sample of 0.150 M acetic acid CH,CO,H is titrated with 0.150 M NaOH soldtion. Calcua pH after the following volumes of base have been added: (a) 0 mL, (b) 17.5 mL, (c) 34.5 mL, (d) 35.0 ml (e) 35.5 mL, (f) 50.0 mL
A student titrated a 100.0 mL sample of 0.100 M acetic acid with 0.050 M NaOH. (For acetic acid, Ka = 1.8 * 10^-5 at this temperature.) (a) Calculate the initial pH. (b) Calculate the pH after 50.0 mL of NaOH has been added. (c) Determine the volume of added base required to reach the equivalence point. (d) Determine the pH at the equivalence point?
A vinegar (acetic acid) solution of unknown concentration was titrated to the light pink endpoint with the standardized NaOH solution. The weight volume % of the vinegar solution were calculated. Molecular formula of Acetic acid: C2H4O2 Volume of vinegar sample titrated (ml) 5.00 Volume of NaOH required to neutralize vinegar in (mL) 8.74 Concentration of NaOH in mol/L, 0.1979 Calculate the weight/volume percentage of the vinegar solution (g/100 ml).
4. A solution of malonic acid, H,CH,04, was standardized by titration with 0.1000 M NaOH solution. If 20.76 mL of the NaOH solution is required to neutralize completely 12.95 mL of the malonic acid solu- tion, what is the molarity of the malonic acid solution? H,CH,O4 +2NaOH- Na C3H,04 +2H,0
25 mL of 0.10 M acetic acid is titrated with 0.10 M NaOH. What is the pH after 30 ml of NaOH have been added? Ka for acetic acid = 1.8 x 10^-5.
40.0 ml of an acetic acid of unknown concentration is titrated with 0.100 M NaOH. After 20.0 mL of the base solution has been added, the pH in the titration flask is 5.10. What was the concentration of the original acetic acid solution? [Ka(CH3COOH) = 1.8 × 10–5]
A solution of 100. ml of .500 M Acetic Acid is titrated with .500 M sodium hydroxide. The Ka of acetic acid is 1.8*10^-5. Find the pH values at the given stages: a) before the addition of any NaOH. B) After 25.0 mL of NaOH added. C) At the equivalence point.
A 25.0 mL sample of 0.293 M acetic acid solution is titrated with 37.5 mL of 0.195 M NaOH solution. What is the pH of this solution if K = 1.8 x 10-5? Final Answer: _______