Mixing 0.02M HCl 100ml with 0.20M CH3COOH (Ka=1.8*10^-5) 100ml obtained a 200ml solution.
Which choose the correct alternative.
a) [H+]=0.11M
b) [H+]=0.01M
c) [CH3COO-]=1.34*10^-3M
d) [Cl-]=0.2M
Mixing 0.02M HCl 100ml with 0.20M CH3COOH (Ka=1.8*10^-5) 100ml obtained a 200ml solution. Which choose the...
The following equilibrium is established in water solution: CH3COOH + H20 = CH3C00+H307 Ka=1.8 x 10-5 Which of the following is false? a CH3COO is the conjugate acid of CH3COOH. b. Hydroxide concentration is negligible with respect to hydrogen ion concentration. OO H2O is a weak base. Od. H2O is the conjugate base of H307 De CH3COOH is a weak acid.
24. 50.00 mL of 0.10 M CH3COOH (Ka = 1.8 x 10-5) is titrated with a 0.10 M KOH solution. After 50.00 mL of the KOH solution is added, the pH in the titration flask will be A) 4.28 B) 8.72 C) 9.41 D) 11.24 E) 12.08 [Kb = 5.6 x 10-10 for acetate ion] 0.0025 - 0.25 .
Consider the titration of 50 mL of 0.100 M acetic acid (CH3COOH, Ka = 1.8 x 10-5) with 0.200 M NaOH solution. Show all calculations for full credit. a) Write the titration reaction: b) Calculate the pH after 5.00 mL of NaOH: c) Calculate the pH after 12.5 mL of NaOH: d) Calculate the pH after 25 mL of NaOH:
1. Calculate the equilibrium concentrations of all species present and the pH of a solution obtained by adding 0.100 moles of solid NaOH to 1.00 L of 15.0 M NH3. Kb = 1.8 × 10–5 2. One mole of a weak acid HA was dissolved in 2.0 L of water. After the system had come to equilibrium, the concentration of HA was found to be 0.45 M. Calculate the Ka for this weak acid. 3. Calculate the pH of a...
John Hogan Quiz 4 LSU Chemistry 1202. Please go into detail on how each answer is obtained as this will be used as a study guide. I will make sure to upvote. We were unable to transcribe this image3. + -10 points 0/100 Submissions Used My Notes If pOH = 5 which of the choices is correct? a) [OH-] = 1.00e-02 M 0 b) pH = 9 c) [OH] = 1.40e-08 M d) [H+] = 1.00e-05 M e) (H+] =...