Question

A 0.551 L solution of 1.81 M oxalic acid (Ka1 = 6.5e-2 and Ka2 = 6.1e-5)...

A 0.551 L solution of 1.81 M oxalic acid (Ka1 = 6.5e-2 and Ka2 = 6.1e-5) is titrated with 1.85 M KOH. What will the pH of the solution be when 0.9326 L of the KOH has been added?

Science   Chemistry   
0 0
Add a commentTranscribed image text
Next >

Homework Answers

Answer #1

0 0
Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Log In

Sign Up

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
A 0.351 L solution of 1.89 M ascorbic acid (Ka1 = 7.9e-5 and Ka2 = 1.6e-12)...
A 0.351 L solution of 1.89 M ascorbic acid (Ka1 = 7.9e-5 and Ka2 = 1.6e-12) is titrated with 1.85 M KOH. What will the pH of the solution be when 0.6419 L of the KOH has been added?
A 0.251 L solution of 1.57 M sulfurous acid (Ka1 = 1.5e-2 and Ka2 = 1.0e-7)...
A 0.251 L solution of 1.57 M sulfurous acid (Ka1 = 1.5e-2 and Ka2 = 1.0e-7) is titrated with 1.85 M NaOH. What will the pH of the solution be when 0.3600 L of the NaOH has been added?
A 0.351 L solution of 1.31 M ascorbic acid (Ka1 = 7.9e-5 and Ka2 = 1.6e-12)...
A 0.351 L solution of 1.31 M ascorbic acid (Ka1 = 7.9e-5 and Ka2 = 1.6e-12) is titrated with 1.65 M NaOH. What will the pH of the solution be when 0.4933 L of the NaOH has been added?
Oxalic acid, found in the leaves of rhubarb and other plants, is a diprotic acid. H2C2O4...
Oxalic acid, found in the leaves of rhubarb and other plants, is a diprotic acid. H2C2O4 + H2O ↔ H3O+ + HC2O4- Ka1= ? HC2O4- + H2O ↔ H3O+ + C2O42- Ka2 = ? An aqueous solution that is 1.05 M H2C2O4 has pH = 0.67. The free oxalate ion concentration in this solution is [C2O42-] = 5.3 x 10-5 M. Determine Ka1 and Ka2 for oxalic acid.
16.5 mL of 0.168 M diprotic acid (H2A) was titrated with 0.118 M KOH. The acid...
16.5 mL of 0.168 M diprotic acid (H2A) was titrated with 0.118 M KOH. The acid ionization constants for the acid are Ka1=5.2×10−5 and Ka2=3.4×10−10. At what added volume (mL) of base does the first equivalence point occur? 20.8 mL of 0.146 M diprotic acid (H2A) was titrated with 0.14 M KOH. The acid ionization constants for the acid are Ka1=5.2×10−5 and Ka2=3.4×10−10. At what added volume (mL) of base does the second equivalence point occur? Consider the titration of...
Oxalic acid, which is formed either as (COOH)2 or H2C2O4, is a diprotic acid. At 25...
Oxalic acid, which is formed either as (COOH)2 or H2C2O4, is a diprotic acid. At 25 degrees celecius ka1= 5.9 x 10^-2 and ka2= 6.4 x 10^-5. Using just sodium oxalate and stock solutions of either 3.00 M HCL (aq), how many grams of sodium oxalate and milliliters of stock solution are needed to prepare 1.00 L of a buffer that has pH= 4.0 and contains 0.16 M oxalate ion?
Calculate the concentrations of H2C2O4, HC2O4−, C2O42−, and H+ ions in a 0.193 M oxalic acid...
Calculate the concentrations of H2C2O4, HC2O4−, C2O42−, and H+ ions in a 0.193 M oxalic acid solution at 25 °C. (Ka1 = 6.5 × 10−2 and Ka2 = 6.1 ×10−5 for oxalic acid.) (Enter your answer in scientific notation.)
An 80.9 mL sample of a 0.100 M solution of a diprotic acid, H2A, is titrated...
An 80.9 mL sample of a 0.100 M solution of a diprotic acid, H2A, is titrated with 0.100 M KOH. For H2A, Ka1 = 8.09 x 10^-5 and Ka2 = 8.09 x 10^-10. A) Calculate the concentration of A2- prior to the addition of any KOH. B) What is the pH of the initial solution? C) What is the pH exactly halfway to the first equivalence point? D) What is the pH exactly halfway between the first and second equivalence...
Calculate the pH at the second stoichiometric point when 160 mL of a 0.020 M solution...
Calculate the pH at the second stoichiometric point when 160 mL of a 0.020 M solution of oxalic acid (Ka1=5.4*10-2, Ka2=5.4*10-5) is titrated with 2.0 M NaOH. What is the most suitable indicator for this titration? bromocresol purple bromocresol green phenol red methyl red phenolphthalein methyl orange t hymol blue thymolphthalein
For the diprotic weak acid H2A, Ka1 = 2.6 × 10-6 and Ka2 = 8.6 ×...
For the diprotic weak acid H2A, Ka1 = 2.6 × 10-6 and Ka2 = 8.6 × 10-9. What is the pH of a 0.0600 M solution of H2A? What are the equilibrium concentrations of H2A and A2– in this solution? pH? H2A? A^2-?
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT