Concept explainers
(a)
Interpretation:
Concept Introduction:
(b)
Interpretation:
Concept Introduction:
(c)
Interpretation:
Concept Introduction:
(d)
Interpretation:
Concept Introduction:
(e)
Interpretation:
Concept Introduction:
The negative log of base 10
Equilibrium constant for base dissociation in aqueous solution is,
The general base dissociation reaction in water is,
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General Chemistry: Atoms First
- Acrylic acid is used in the polymer industry in the production of acrylates. Its K, is 5.6 X 10“’. What is the pH of a 0.11 M solution of acrylic acid, CH2CHCOOH?arrow_forwardThe 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.)arrow_forwardA 30.0 mL sample of 0.135 M aqueous benzoic acid is titrated with 0.115 M aqueous potassium hydroxide. (a) What is the initial pH of the benzoic acid solution? (b) What is the pH after the addition of 5.00 mL potassium hydroxide solution? (c) What volume of potassium hydroxide solution is required to reach halfway to the equivalence point? (d) Calculate the pH at the point halfway to the equivalence point. (e) What volume of potassium hydroxide solution is required to reach the equivalence point? (see [c]) (f) Calculate the pH at the equivalence point. (g) Calculate the pH of the solution when an additional 5.0 mL of potassium hydroxide has been added (5.00 mL in addition to the amount required to reach the equivalence point).arrow_forward
- In the titration of 60.0 mL of 1.0 M methylamine, CH3NH₂ (K₁ = 4.4 x 10-4), with 0.50 M HCI, calculate the pH under the following conditions. (a) after 0.00 mL of 0.50 M HCI has been added 4.0 (b) after 20.0 mL of 0.50 M HCI has been added 4.0✔ (c) after 60.0 mL of 0.50 M HCI has been added 4.0✔ (d) at the stoichiometric point 4.0✔ (e) after 200.00 mL of 0.50 M HCI has been added 4.0arrow_forward(7) Calculate the pH of each of the following solutions: (a) 0.1000M Propanoic acid( HC H O,,K=1.3x105) (b) 0.1000M sodium propanoate (Na C HỎ) (c) 0.1000M HC₂H₂O, and 0.1000M Nа С¸¸0₂ 3 5 52 (d) After 0.020 mol of HCl is added to 1.00 L solution of (a) and (b) above. (e) After 0.020 mol of NaOH is added to 1.00 L solution of (a) and (b) above.arrow_forwardDetermine the pH during the titration of 67.3 mL of 0.459 M hypochlorous acid (K 3.5x108) by 0.459 M NaOH at the following points. (Assume the titration is done at 25 °C.) (a) Before the addition of any NaOH X (b) After the addition of 17.0 mL of NaOH X t (c) At the half-equivalence point (the titration midpoint) (d) At the equivalence point x pt (e) After the addition of 101 mL of NaOH 12.95 pt pt Xarrow_forward
- (a) Calculate the pH of the 0.30 M NH3 / 0.35 M NH4Cl buffer. What is the pH of the buffer after the addition of 0.030 mol HCl? note: Ka (NH4+) = 5.6 x 10 -10 NH3 (aq) + H+ (aq) → NH4+ (aq) (b) What are the hydronium [H3O+] and hydroxide [OH-] ion concentrations at 25°C in a 4.0 M aqueous Mg(OH)2.arrow_forward4) A highly toxic hydrogen cyanide (HCN) is a weak acid. A chemical engineer plans to determine pH of a 50 mL sample of HCN (0.10 M) in a titration process. To this end, she used 0.20 M NaOH as a titrant in varying volumes. Calculate the pH of the solution at the following points: (Ka for HCN=6.2×10-¹0) (a) Before addition of NaOH (initial pH), (b) After 10.00 mL of titrant addition, (c) After 25.00 mL of titrant addition, (d) After 50.00 mL of titrant addition.arrow_forwardA buffer solution was prepared that contained 0.60 M hydrogen fluoride, HF (Ka = 7.2 x 104) and 1.00M potassium fluoride, KF. The total volume was 250 mL. (a) What ions and molecules are present in the solution? List them in order of decreasing concentration: Decreasing order of Concentration (b) What is the pH of the buffer solution described above? (c) What is the pH of 100. mL of the buffer solution if you add 100. x 10-3 g of NaOH? Assume negligible change in volume. (USEFUL INFORMATION: MM NaOH = 39.997 g mol-1)arrow_forward
- Consider the titration of a weak acid that has a pKa = 4.00. Suppose a chemist was going to perform a titration on 50.0 mL of 0.050 0 M of the weak acid using 0.500 M NaOH. (A) What would be the pH of the solution after 0.00 mL of 0.500 M NaOH has been added?(B) What would be the pH halfway to the equivalence point of the titration?(C) What would be the pH at the equivalence point in the titration? (D) What would be the pH of the solution after 6.00 mL of 0.500 M NaOH has been added? Use the following reasoning when solving this problem. Because at the equivalence point the moles of strong base (0.500 M NaOH) added as a titrant equal the moles of weak acid, HA (50.0 ml of 0.050 M) in the solution, we can start by calculating the volume of 0.500 M NaOH needed to reach the equivalence point. To do this we simply equate the moles of HA to moles of NaOH by using the dilution formula. i.e. MaVa = MbVb where a represents HA and b represents NaOH. To calculate volume of NaOH needed to reach…arrow_forwardYou prepared 500 mL buffer solution that contains 0.50 moles monoprotic weak acid (HA) and 0.30 moles conjugate base (NaA). The solution has a pH of 4.20. (a) What is the Kp of the conjugate base, NaA? (b) What is the pH of the buffer solution after adding 400 mL of 0.30 M H,SO4? pH = pKa – log(THA), [A*]arrow_forwardFor each of the following cases, decide whether the pH is less than 7, equal to 7, or greater than 7. (a) Equal volumes of 0.10 M acetic acid, CH3CO2H, and 0.10 M KOH are mixed. (b) 25 mL of 0.015 M NH3 is mixed with 25 mL of 0.015 M HCl. (c) 150 mL of 0.20 M HNO3 is mixed with 75 mL of 0.40 M NaOH.arrow_forward
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