Question

In: Chemistry

Lab inspired question: A student titrated 10.00mL of a solution of monoprotic lactic acid (CH3H5O3H)solution pKa=...

Lab inspired question:

A student titrated 10.00mL of a solution of monoprotic lactic acid (CH3H5O3H)solution pKa= 3.86 with 0.2M sodium hydroxide.

It required 5.00mL of 0.2M NaOH to completely neutralize the lactic acid solution.

Calculate the concentration of the unknown lactic acid solution in mol/L.

The student attempted to titrate 10.00mL of a solution of oxalic acid, a diprotic acid, H2C2O4 with the same concentration as the lactic acid solution from the earlier question. What volume of 0.2M NaOH would be required to reach the equivalence point?

Solutions

Expert Solution


Related Solutions

Lactic acid (HC3H5O3), a weak monoprotic acid with a Ka = 1.4 x 10-4, is titrated...
Lactic acid (HC3H5O3), a weak monoprotic acid with a Ka = 1.4 x 10-4, is titrated with KOH. If you have 50.0 mL of a 0.500 M HC3H5O3solution, calculate the pH after 150.0 mL of 0.250 M KOH have been added:
1)Lactic acid (HC3H5O3), a weak monoprotic acid with a Ka = 1.4 x 10-4, is titrated...
1)Lactic acid (HC3H5O3), a weak monoprotic acid with a Ka = 1.4 x 10-4, is titrated with KOH. If you have 50.0 mL of a 0.500 M HC3H5O3 solution, calculate the pH after 100.0 mL of 0.250 M KOH have been added: 2)Lactic acid (HC3H5O3), a weak monoprotic acid with a Ka = 1.4 x 10-4, is titrated with KOH. If you have 50.0 mL of a 0.500 M HC3H5O3solution, calculate the pH after 20.0 mL of 0.250 M KOH...
A weak acid was HA (pKa=5.00) was titrated with 1.00M KOH. The acid solution had a...
A weak acid was HA (pKa=5.00) was titrated with 1.00M KOH. The acid solution had a volume of 100ml and a molarity of 0.100M. Find the pH at the following volumes of base added Vb= 0, 1, 5, 9, 9.9, 10, 10.1 and 12.
A weak acid HA (pKa = 5.00) was titrated with 1.10 M KOH. The acid solution...
A weak acid HA (pKa = 5.00) was titrated with 1.10 M KOH. The acid solution had a volume of 100.0 mL and a molarity of 0.107 M. Find the pH at the following volumes of base added: Vb = 0.00, 1.00, 5.00, 9.00, 9.90, 10.00, 10.10, and 12.00 mL. (Assume Kw = 1.01 ✕ 10−14.)
the pka of hypochlorous acid is 7.530. A 60 ml solution of .102M NaCl is titrated...
the pka of hypochlorous acid is 7.530. A 60 ml solution of .102M NaCl is titrated with .286 M Hcl. calculate the ph of the solution after 1) after the addition of 6.87ml of .286M Hcl 2)After the addition of 22.7ml of .286 M Hcl 3) at the equivalence point with .286 M Hcl
A 35. mL of a 0.045 M solution of acetic acid, pKa = 4.74, is titrated...
A 35. mL of a 0.045 M solution of acetic acid, pKa = 4.74, is titrated with a 0.068 M solution of NaOH. What is the hydroxide ion concentration of the resulting solution after 30.0 mL of NaOH is added?
A 10.00 mL portion of a solution of acetic acid (pKa = 4.7) is titrated against...
A 10.00 mL portion of a solution of acetic acid (pKa = 4.7) is titrated against a standardized solution of 0.200 M NaOH. The equivalence point is observed after 30.00 mL of base have been added. After which total volume of added base will the pH be closest to 4.7? A. 0.00 mL b. 5.00 mL c. 10.00 mL d. 15.00 mL e. 20.00 mL f. 30.00 mL -Please explain, Thank you.
1) 20.00 mL of a 0.3000 M lactic acid solution is titrated with 0.1500 M NaOH....
1) 20.00 mL of a 0.3000 M lactic acid solution is titrated with 0.1500 M NaOH. a. What is the pH of the initial solution (before any base is added)? b. What is the pH of the solution after 20.00 mL of the base solution has been added? c. What is the pH of the solution after 40.00 mL of the base solution has been added?
50mL of a 0.1000 M aqueous solution of hydrazoic acid (HN3, pKa=4.72) is titrated with a...
50mL of a 0.1000 M aqueous solution of hydrazoic acid (HN3, pKa=4.72) is titrated with a 0.1000M solution of sodium Hydroxide. Calculate the pH at the following points a. Before Titration b. After the addition of 5.00mL of NaOH c. After the addition of 25.00mL of NaOH d. After the addition of 50.00mL of NaOH e. After the addition of 51.00mL of NaOH. PLEASE EXPLAIN STEPS AND NOT JUST SHOW WORK, I KNOW THE FIRST ONE, THE OTHERS ARE GIVING...
50mL of a 0.1000 M aqueous solution of hydrazoic acid (HN3, pKa=4.72) is titrated with a...
50mL of a 0.1000 M aqueous solution of hydrazoic acid (HN3, pKa=4.72) is titrated with a 0.1000M solution of sodium Hydroxide. Calculate the pH at the following points a. Before Titration b. After the addition of 5.00mL of NaOH c. After the addition of 25.00mL of NaOH d. After the addition of 50.00mL of NaOH e. After the addition of 51.00mL of NaOH. PLEASE EXPLAIN STEPS AND NOT JUST SHOW WORK, I KNOW THE FIRST ONE, THE OTHERS ARE GIVING...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT