calculate the pH and pOH for a 0.15 M oxalic acid H2C2O4 solution.
(ka1-6.0x10^-2, Ka2-6.1x10^-5)
calculate the pH and pOH for a 0.15 M oxalic acid H2C2O4 solution. (ka1-6.0x10^-2, Ka2-6.1x10^-5)
Calculate the pH and the equilibrium concentrations of HC2O4- and C2O42- in a 0.0782 M oxalic acid solution, H2C2O4 (aq). For H2C2O4, Ka1 = 5.9×10-2 and Ka2 = 6.4×10-5 pH =_______ [HC2O4-] =___________ M [C2O42-] =___________ M
Calculate the pH and the equilibrium concentrations of HC2O4- and C2O42- in a 0.1160 M oxalic acid solution, H2C2O4 (aq). For H2C2O4, Ka1 = 5.9×10-2 and Ka2 = 6.4×10-5 pH = [HC2O4-] = M [C2O42-] = M
Calculate the pH of 1.00 x 10^-2 M oxalic acid solution. Calculate the pH of 1 x 10^-2 M solution of sodium oxalate. Ka1 = 5.62 x 10^-2, Ka2 = 5.42 x 10^-5
Determine the concentration of H3O+ in a 0.064 M solution of oxalic acid, H2C2O4.H2C2O4(aq)+H2O(l) ⇌ H3O+(aq)+HC2O−4(aq) Ka1=0.054HC2O−4(aq)+H2O(l) ⇌ H3O+(aq)+C2O2−4(aq) Ka2=5.4×10−5
QUESTION: Calculate the pH of a 0.05 M solution of ascorbic acid (Ka1 = 7.9x10–5; Ka2 = 1.6x10–12). SHOW YOUR WORK STEP BY STEP
A 27 mL sample of 0.100 M oxalic acid (Ka1= 5.60 * 10^-2, Ka2= 5.42* 10^-5) was titrated with 0.0850 M NaOH. What is the pH at 4 mL before the equivalence point? What is the pH at 4 mL after the equivalence point?
The pH of a 0.20-M solution of oxalic acid (H2C2O4) is measured to be 1.10. Use this information to determine a value of Ka for oxalic acid. H2C2O4(aq) + H2O(l) HC2O4-(aq) + H3O+(aq) Ka =
A 27 mL sample of 0.100 M oxalic acid (Ka1= 5.60 * 10^-2, Ka2= 5.42* 10^-5) was titrated with 0.0850 M NaOH. What is the pH at 4 mL before the equivalence point? What is the pH at 4 mL after the equivalence point?
Oxalic acid, H2C2O4 has acid dissociation constants of ?a1=5.90×10−2 and ?a2=6.40×10−5. Calculate the pH and molar concentrations of H2C2O4 , HC2O−4 , and C2O2−4 at equilibrium for each of the solutions. A 0.117 M solution of H2C2O4 A 0.117 M solution of Na2C2O4
Consider malonic acid, H2C3H2O4. Calculate the pH, pOH, [H2C3H2O4], [HC3H2O4 -], and [C3H2O4 2-] for a 0.750 M solution of malonic acid. Ka1 for Malonic Acid = 1.4 x 10-4 Ka2 for Malonic Acid = 2.0 x 10