Question

In: Chemistry

A 0.02500 g sample of the diprotic acid malonic acid (H2C3H2O4) is dissolved in water to...

A 0.02500 g sample of the diprotic acid malonic acid (H2C3H2O4) is dissolved in water to yield a 2.00 L solution. What is the resulting pH of the solution and the concentration of malonate ion (C3H2O4^2-) at equilibruim?

Solutions

Expert Solution


Related Solutions

Malonic acid (H2M) is a diprotic acid. You wish to prepare a buffer of malonic acid...
Malonic acid (H2M) is a diprotic acid. You wish to prepare a buffer of malonic acid (pKa1 = 2.847, pKa2= 5.696) with a final pH of 2.90 but you only have disodium malonate (M2-) in your laboratory shelf. You dissolve 100 mmol of this salt in 1 L of water. Assume both dissociation processes are decoupled. What is the initial pH of the solution? State any assumptions you make. How many equivalents of strong acid must you add to your...
A 425 mg sample of a weak diprotic acid is dissolved in enough water to make...
A 425 mg sample of a weak diprotic acid is dissolved in enough water to make 275.0 ml of solution. The pH of this solution is 3.08. A saturated solution of calcium hydroxide (Ksp= 1.3 x 10-6) is prepared by adding excess calcium hydroxide to water and then removing the undissolved solid by filtration. Enough of the calcium hydroxide solution is added to the solution of the acid to reach the second equivalence point. The pH at the second equivalence...
A 425 mg sample of a weak diprotic acid is dissolved in enough water to make...
A 425 mg sample of a weak diprotic acid is dissolved in enough water to make 275.0 ml of solution. The pH of this solution is 3.08. A saturated solution of calcium hydroxide (Ksp= 1.3 x 10^-6) is prepared by adding excess calcium hydroxide to water and then removing the undissolved solid by filtration. Enough of the calcium hydroxide solution is added to the solution of the acid to reach the second equivalence point. The pH at the second equivalence...
A 40.00 mL sample of 0.1000 M diprotic malonic acid is titrated with 0.0900 M KOH....
A 40.00 mL sample of 0.1000 M diprotic malonic acid is titrated with 0.0900 M KOH. What volume KOH must be added to give a pH of 6.00 ? Ka1 = 1.42 × 10−3 and Ka2 = 2.01 × 10−6.
A 25.00 mL sample of 0.120 M of the diprotic malonic acid, HOOC-CH2-COOH, was titrated with...
A 25.00 mL sample of 0.120 M of the diprotic malonic acid, HOOC-CH2-COOH, was titrated with 0.250 M NaOH. If the following results were obtained, calculate Ka1 and Ka2. mL NaOH added: 5.00 6.00 10.00 12.00 15.00 18.00 20.00 24.00 pH: 2.68 2.83 3.53 4.26 5.21 5.69 6.00 9.24
A 0.1276 g sample of an unknown monoprotic acid was dissolved in 25.0 mL of water...
A 0.1276 g sample of an unknown monoprotic acid was dissolved in 25.0 mL of water and titrated with 0.0633 M NaOH solution. The volume of base required to bring the solution to the equivalence point was 18.4 mL. (a) Calculate the molar mass of the acid. (b) After 10.0 mL of base had been added during the titration, the pH was determined to be 5.87. What is the Ka of the unknown acid? (10 points)
A 6.31 g sample of benzoic acid was dissolved in water to give 56.5 mL of...
A 6.31 g sample of benzoic acid was dissolved in water to give 56.5 mL of solution. This solution was titrated with 0.260 M NaOH . What was the pH of the solution when the equivalence point was reached? Ka of benzoic acid is 6.3x10 to the negative fifth .
A 0.520 g sample of a diprotic acid with a molar mass of 255.8 g/mol is...
A 0.520 g sample of a diprotic acid with a molar mass of 255.8 g/mol is dissolved in water to a total volume of 23.0 mL . The solution is then titrated with a saturated calcium hydroxide solution. a. Assuming that the pKa values for each ionization step are sufficiently different to see two equivalence points, determine the volume of added base for the first and second equivalence points. b. The pH after adding 23.0 mL of the base was...
You have a 1.153 g sample of an unknown solid acid, HA, dissolved in enough water...
You have a 1.153 g sample of an unknown solid acid, HA, dissolved in enough water to make 20.00 mL of solution. HA reacts with KOH(aq) according to the following balanced chemical equation: HA(aq)+KOH(aq)--->KA(aq)+H2O(l) if 13.40 mL of 0.715 M KOH is required to titrate the unknown acid to the equivalence point, what is the concentration of the unknown acid? ALSO What is the molar mass of HA?
A 0.1219 g sample of a Vitamin C (ascorbic acid) tablet was dissolved in acid. A...
A 0.1219 g sample of a Vitamin C (ascorbic acid) tablet was dissolved in acid. A 25.00 mL aliquot of 0.01739 M KIO3 was added along with excess KI. The resulting solution was titrated with 0.07211 M thiosulfate, requiring 21.44 mL to reach the endpoint. Compute the weight percent of ascorbic acid (FW = 176.12) in the tablet.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT