A 100.0 mL solution containing 0.976 g of maleic acid (MW = 116.072 g/mol) is titrated...

A 100.0 mL solution containing 0.976 g of maleic acid (MW = 116.072 g/mol) is titrated with 0.296 M KOH. Calculate the pH of the solution after the addition of 56.00 mL of the KOH solution. Maleic acid has pKa values of 1.92 and 6.27

. pH =

At this pH, calculate the concentration of each form of maleic acid in the solution at equilibrium. The three forms of maleic acid are abbreviated H2M, HM–, and M2–, which represent the fully protonated, intermediate, and fully deprotonated forms, respectively.

[M2-] = ___ M

[HM-] = ___ M

[H2M] = ___ M

Expert Solution

queen_honey_blossom answered 1 year ago

A 100.0 mL solution containing 0.830 g of maleic acid (MW = 116.072 g/mol) is titrated...

A 100.0 mL solution containing 0.830 g of maleic acid (MW = 116.072 g/mol) is titrated with 0.298 M KOH. Calculate the pH of the solution after the addition of 48.0 mL of the KOH solution. Maleic acid has pKa values of 1.92 and 6.27. pH = At this pH, calculate the concentration of each form of maleic acid in the solution at equilibrium. The three forms of maleic acid are abbreviated H2M, HM–, and M2–, which represent the fully...

A 100.0 mL solution containing 0.8366 g of maleic acid (MW = 116.072 g/mol) is titrated...

A 100.0 mL solution containing 0.8366 g of maleic acid (MW = 116.072 g/mol) is titrated with 0.3203 M KOH. Calculate the pH of the solution after the addition of 45.00 mL of the KOH solution. Maleic acid has pKa values of 1.92 and 6.27. At this pH, calculate the concentration of each form of maleic acid in the solution at equilibrium. The three forms of maleic acid are abbreviated H2M, HM–, and M2–, which represent the fully protonated, intermediate,...

A 100.0 mL solution containing 0.7590 g of maleic acid (MW = 116.072 g/mol) is titrated...

A 100.0 mL solution containing 0.7590 g of maleic acid (MW = 116.072 g/mol) is titrated with 0.2906 M KOH. Calculate the pH of the solution after the addition of 45.00 mL of the KOH solution. Maleic acid has pKa values of 1.92 and 6.27. pH=? At this pH, calculate the concentration of each form of maleic acid in the solution at equilibrium. The three forms of maleic acid are abbreviated H2M, HM-, and M^2-, which represent the fully protonated,...

A 100.0 mL solution containing 0.753 g of maleic acid (MW = 116.072 g/mol) is titrated...

A 100.0 mL solution containing 0.753 g of maleic acid (MW = 116.072 g/mol) is titrated with 0.259 M KOH. Calculate the pH of the solution after the addition of 50.0 mL of the KOH solution. Maleic acid has pKa values of 1.92 and 6.27. At this pH, calculate the concentration of each form of maleic acid in the solution at equilibrium. The three forms of maleic acid are abbreviated H2M, HM–, and M2–, which represent the fully protonated, intermediate,...

A 100.0 mL solution containing 0.756 g of maleic acid (MW = 116.072 g/mol) is titrated...

A 100.0 mL solution containing 0.756 g of maleic acid (MW = 116.072 g/mol) is titrated with 0.217 M KOH. Calculate the pH of the solution after the addition of 60.0 mL of the KOH solution. Maleic acid has pKa values of 1.92 and 6.27. At this pH (pH = 9.48), calculate the concentration of each form of maleic acid in the solution at equilibrium. The three forms of maleic acid are abbreviated H2M, HM–, and M2–, which represent the...

An aqueous solution containing 1.00 g of oxobutanedioc acid (FM 132.07) per 100 mL was titrated...

An aqueous solution containing 1.00 g of oxobutanedioc acid (FM 132.07) per 100 mL was titrated with 0.09432 M NaOH. Calculate the pH at th following volumes of added base: 0.5Ve1, Ve1, 1.5Ve2, Ve2, 1.05Ve2 Sketch the titration curve, using the values calculated above. Which equivalence point would be best to use in this titration (which one is not blurred)? Ka1 = 2.56 Ka2 = 4.37 Ve1 = equivalence point 1 Ve2 = equivalence point 2

A solution of 100.0 mL of 1.00 M malonic acid (H2A) was titrated with 1.00 M...

A solution of 100.0 mL of 1.00 M malonic acid (H2A) was titrated with 1.00 M NaOH. K1 = 1.49 x 10-2, K2 = 2.03 x 10-6. What is the pH after 50.00 mL of NaOH has been added?

1) calculate the concentration of maleic acid solution?given : 25.00 mL of maleic acid of unknown...

1) calculate the concentration of maleic acid solution?given : 25.00 mL of maleic acid of unknown contration is diluted with 50.00 mL of water. It is used to titrate 49.72mL of 0.1 M NaOH solution. 2)calculate the concentration of the diluted and undiluted CH3COOH (aq) solution. given: 25.00 mL of unknown acetic acid solution is diluted with 50.00 mL water. It is used to titrate 49.83 mL of 0.1 M NaOH solution

Calculate the pH of each solution. A. A solution containing 0.0133 M maleic acid and 0.0189...

Calculate the pH of each solution. A. A solution containing 0.0133 M maleic acid and 0.0189 M disodium maleate. The Ka values for maleic acid are 1.20×10−2 (Ka1) and 5.37×10−7 (Ka2). B. A solution containing 0.0368 M succinic acid and 0.022 M potassium hydrogen succinate. The Ka values for succinic acid are 6.21×10−5 (Ka1) and 2.31×10−6 (Ka2).

A sample containing a mixture of SrCl2·6H2O (MW = 266.62 g/mol) and CsCl (MW = 168.36...

A sample containing a mixture of SrCl2·6H2O (MW = 266.62 g/mol) and CsCl (MW = 168.36 g/mol) originally weighs 1.7215 g. Upon heating the sample to 320 °C, the waters of hydration are driven off SrCl2·6H2O, leaving the anhydrous SrCl2. After cooling the sample in a desiccator, it has a mass of 1.2521 g. Calculate the weight percent of Sr, Cs, and Cl in the original sample.

Question