Answered: Calculate the pH during the titration…

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter15: Complex Acid/base Systems

Section: Chapter Questions

Problem 15.4QAP

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The hydrogen phthalate ion, C8HsO4, is a weak acid with Ka = 3.91 106. C8H5O4(aq)+H2O(l)C8H4O42(aq)+H3O+(aq) What is the pH of a 0.050 M solution of potassium hydrogen phthalate. KC8H5O4? Note: To find the pH for a solution of the anion, we must take into account that the ion is amphiprotic. It can be shown that, for most cases of amphiprotic ions, the H3O+ concentration is [H3O+]=Ka1Ka2 For phthalic acid, C8H6O4 is Ka1 is 1.12 103, and Ka2 is 3.91 106.
Assume you dissolve 0.235 g of the weak acid benzoic acid, C6H5CO2H, in enough water to make 1.00 102 mL of solution and then titrate the solution with 0.108 M NaOH. C6H5CO2H(aq) + OH(aq) C6H5CO2(aq) + H2O() (a) What was the pH of the original benzoic add solution? (b) What are the concentrations of all of the following ions at the equivalence point: Na+, H3O+, OH, and C6H5CO2? (c) What is the pH of the solution at the equivalence point?
The simplest amino acid is glycine, H2NCH2CO2H. The common feature of amino acids is that they contain the functional groups: an amine group, -NH2, and a carboxylic acid group, -CO2H. An amino acid can function as either an acid or a base. For glycine, the acid strength of the carboxyl group is about the same as that of acetic acid. CH3CO2H, and the base strength of the amino group is slightly greater than that of ammonia, NH3. (a) Write the Lewis structures of the ions that form when glycine is dissolved in 1 M HCl and in 1 M KOH. (b) Write the Lewis structure of glycine when this amino acid is dissolved in water. (Hint: Consider the relative base strengths of the -NH2 and -CO2- groups.)
The text describes zinc hydroxide as an amphoteric hydroxide, so Zn(OH)2 is soluble in basic solution. What is the molar solubility of Zn(OH)2 in 1.00 M NaOH? What is the pH of the equilibrium solution?
The student titrated 50.0 mL of 0.100 M HC2H3O2 with 0.100 M NaOH (aq). Calculate the pH of the solution at the equivalence point.
Calculate the pH during the titration of 20.00 mL of 0.1000 M CH3CH2HOOH(aq) with 0.1000 M NaOH(aq) after 21.33 mL of the base have been added.
Calculate the pH of the resulting solution if 35.0 mL0.350 M HCl(aq) is added to 25.0 mL of 0.450 M NaOH(aq).
A solution is prepared by adding 100 mL of 1.0 M HC,H,O,(aq) to 100 mL of 1.0 M NaC,H,O,(aq). The solution is stirred and its pH is measured to be 4.73. After 3 drops of 1.0 M HCl are added to the solution, the ph of the solution is measured and is still 4.73. Which of the following equations represents the chemical reaction that accounts for the fact that acid was added but there was no detectable change in pH? (A) H;O*(aq) + OH (aq) → 2 H,O(1) (B) H;O*(aq) + CI (aq) → HCI(g) + H,O(1) (C) H,O*(aq) + C,H,O, (aq) → HC,H,0,(aq) + H,0(1) (D) H;O*(aq) + HC,H,O,(aq) - H,C,H,0,*(aq) + H,O()
2.500 g of an unknown achiral primary amine is dissolved in enough water to yield a 50.00 mL solution. This solution is titrated with 0.9250 M HCl(aq). After 24.95 mL of titrant is added, the pH of the resulting solution is 10.28. After a total of 36.95 mL of titrant is added, the equivalence point is reached. Calculate the molar mass (MM) of the unknown amine.
Calculate the change in pH when 5.00 mL of 0.100 M HCl(aq) is added to 100.0 mL of a buffer solution that is 0.100 M in NH3(aq) and 0.100 M in NH4Cl(aq). A list of ionization constants can be found here. B)Calculate the change in pH when 5.00 mL of 0.100 M NaOH(aq) is added to the original buffer solution.
Determine the pH at the equivalence (stoichiometric) point in the titration of 48.38 mL of 0.171 M HCN (aq) with 0.135 M NaOH(aq). The Ka of HCN is 4.9 * 10^-10.
in the titration of 50.0 mL of 0.0200 M C6H5COOH(aq) with 0.100 M NaOH(aq), what is/are the major species in the solution after the addition of 5.0 mL of NaOH(aq)?

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Calculate the pH during the titration of 20.0 mL of 0.30 M HNO3(aq) with 0.30 M KOH after 19.24 mL of the base have been added.