50.0 mL of 0.100 M KHC8H4O4 + 25.0 mL of 0.200 M NaOH. (pH at equivalent point)

Question

50.0 mL of 0.100 M KHC8H4O4 + 25.0 mL of 0.200 M NaOH. (pH at equivalent point) When 10.0 mL solution of 0.100 M NaOH is added to 40.0 mL of 0.100 M acrylic acid (HC3H3O2) the resulting solution is found to have pH - 3.78 (a) Write the net ionic equation for the reaction between hydroxide ion and acrylic acid and determine the Ka of acrylic acid. Calculate the Kb of the acrylate ion, C3H3Q2 (conjugate base), and the pH of a solution containing 0.050 M sodium acrylate, NaC3H3O2. In an acid-base titration experiment, 50.0 mL of 0 100 M potassium hydrogen phthalate (KHCgH4O4) is titrated with 0.200 M NaOH solution. The net ionic equation for the titration reaction is as follows. HC8H4O4- (aq) + OH- (aq) rightarrow C8H4O42-(aq) + H2O(l) If hydrogen phthalate ion, HC8H4O4- has Ka = 4.0 x 10-6, calculate the pH of each of the following solutions: (a) Starting pH for 0.100 M KHC8H4O4 (before NaOH solution is added). 50.0 mL of 0.100 M KHC8H4O4 + 10.0 mL of 0.200 M NaOH; 50.0 mL of 0.100 M KHC8H4O4 + 20.0 mL of 0.200 M NaOH;

Explanation / Answer

I think you are doing it the correct way.

Related Questions

Navigate

• Browse (All)
• Browse 5
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.