Ascorbic acid, H₂C6H6O6, is a diprotic acid usually known as vitamin C. For this acid, pk1 is 4.10 and pKa2 is 11.79. When 125 mL of asolution of ascorbic acid was evaporated to dryness, the residue of pure ascorbic acid had a mass of 3.30 g.Calculate the molar concentration of ascorbic acid in the solution before it was evaporated. iWhat was the pH of the ascorbic acid solution before it was evaporated? pH = iWhat was the concentration of the ascorbate ion, C6H6062, before the solution was evaporated? iMM

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Answered: Ascorbic acid, H₂C6H6O6, is a diprotic…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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You have 1 M (1 mol/L) solution of amino acid. This amino acid has an alpha carboxyl group with pKa = 2.0 and an alpha- amino group with pKa = 10.0. The pH of the solution is 7.0. Calculate the concentration of the molecular form of this amino acid that has carboxyl group protonated (COOH) and, at the same time, the amino group deprotonated (NH2).
Pleas give the structure of the major organic product in the reaction
A weak acid, HA, has a pKa = 6.2. If a solution of NaA(aq) is prepared, what does the pH of the solution need to be adjusted to in order to achieve the following ratios of deprotonated to protonated species? 1) [A-] / [HA] = 0.01 2) [A-] / [HA] = 10 3) [A-] / [HA] = 1 4) [A-] / [HA] = 106 Note: NaA is the sodium salt of , the conjugate base of HA. Assume NaA completely dissociates in water.
M11.
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Alanine (HA) is a diprotic amino acid with Kal = 4.53 x 10-3 and K₁2 = 1.36 x 10-10. Determine the pH of each of the solutions. A 0.241 M alanine hydrochloride (H₂A+CI-) solution. pH = A 0.241 M alanine (HA) solution. pH = A 0.241 M sodium alaninate (Na+ A¯) solution. pH =
Calculate the ratio of dihydrogen phosphate to hydrogen phosphate in blood at pH 7.4. The Ka is 6.3 × 10−8.
What is the equilibrium constant for the following reaction that occurs when disodium citrate(Na2HC6H5O7) dissolves in water at 25ºC? The acid dissociation constants for citric acid, H3C6H5O7 , are:Ka1 = 7.4 × 10–4 , Ka2 =1.7 × 10–5 , and Ka3 = 4.0 × 10–7 . HC6H5O72– (aq) + H2O(l) ⇌ H2C6H5O7– (aq) + OH– (aq)
For an acid HA with a pKa of 5.2, what is the ratio of the concentration of the conjugate base to that of the acid at pH 7.2.
What is the pH of a solution that is a prepared by dissolving 4.90 g of (NH4)2SO4 ( K₁ (NH4+) = 5.70 × 10-¹0) in water, adding 150.0 mL of 0.1263 M NaOH, and diluting to 900.0 mL? pH = b0.124 M in piperidine and 0.054 M in its chloride salt ( Ka (C5H₁1NH+) = 7.50 × 10−¹² )? -12 pH = C prepared by dissolving 2.12 g of aniline (93.13 g/mol, Ką (anilinium ion) = 2.51 × 10−5) in 100. mL of 0.0330 M HCl and diluting to 500.0 mL? pH =
Calculate the pH at 25 °C of a 0.38M solution of trimethylammonium chloride pk of 4.19. Round your answer to 1 decimal place. pH = X Ś ((CH3),NHCI). Note that trimethylamine ((CH3)3 N is a weak base with a

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