Concept explainers
Interpretation:
For which of the given weak acid solution can we most likely neglect x in the denominator of the equilibrium expression in the determination of pH has to be identified
Concept Information:
Weak acid:
Weak acids ionize only to limited extent in water. At equilibrium, an aqueous solution of a weak acid has a mixture of unionized acid molecules, hhydronium ions and corresponding conjugate base.
pH definition:
The concentration of hydrogen ion is measured using
The
To Identify: The diagram representing weak acid for which x can be neglected in the denominator of equilibrium expression
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Chemistry: Atoms First
- What is the pH of the solution obtained by titrating 1.30 g of sodium hydrogen sulfate, NaHSO4, dissolved in 50.0 mL of water with 0.175 M sodium hydroxide until the equivalence point is reached? Assume that any volume change due to adding the sodium hydrogen sulfate or to mixing the solutions is negligible.arrow_forward(a) Calculate the pH of a buffer solution containing 0.40 mole of KF and 0.50 mol of HF in 1.50 liter solution. (Ka of HF is 7.2*10–4); 15.0 mL of 1.00 M HCl is added to the original buffer in question (a). (b) Show the equation of reaction which take place when HCl is added. (c) What will the pH of the buffer solution in question (a) be after the addition of 15.0 mL of 1.00 M HCl.arrow_forwardThree buffers are prepared with equal concentrations of formic acid and sodium formate (Buffer 1), hydrofluoric acid and sodium fluoride (Buffer 2), and acetic acid and sodium acetate (Buffer 3). Rank the three buffers from highest to lowest pH at which they can buffer a solution (i.e., their buffering pH).(a) 1, 2, 3(b) 3, 2, 1(c) 2, 1, 3(d) 3, 1, 2arrow_forward
- Ammonia is a convenient buffer system in the slightly basic range. (a) What is the pH of a buffer solution containing 24 g of NH4Cl dissolved in 1.00 L of 0.879 M NH3? pH = 9.5521 (b) How many moles of acid are required to change the pH of this solution by 0.06 pH units? i 0.23 pH = (c) Suppose 4.8 mL of 12.9 M HCI solution is added to 362 mL of the solution of Part (a). Calculate the new pH. mol i 9.08arrow_forwardA solution NH3 that contains 72 mL of 0.043 M ammonia, NH3, is titrated with 0.083 M HCl. The Kb of ammonia is 1.8x10-5. (a) What volume of 0.083 M HCI would be added to reach the equivalence point? Give the volume in mL. 49 37 mL (b) At the equivalence point, what is the pH of the solution? (Assume that volumes are additive.) 40 8.98 X whawhat "at onubralence point" implies about the quantities of the combined acids and bases? Did you rememberarrow_forward1.56 g of sodium acetate, NaCH;CO, has been 0.20 M ammonia, NH3, and 0.20 M ammonium 4. What is the pH of 0.15 M acetic acid to which 3. What is the pH of a solution that consists of aqueous solution of NH3? the same when you (c) add solid NaCl to a dilute aqueous solution of (b) add solid sodium acetate to a dilute a (a) add solid ammonium chloride to a dilute 1. Does the pH of the solution increase, decrease, or 17.1 and 17.2.) stay solution of acetic acid? aqueous NaOH? 2 Does the pH of the solution increase, decrease, or stay the same when you (a) add solid sodium oxalate, Na,C,O4, to 50.0 mL of 0.015 M oxalic acid, H,C,O4? (b) add solid ammonium chloride to 75 mL of 0.016 M HCl? (c) add 20.0 acetate, NaCH;CO2? of NaCl to 1.0 L of 0.10 M sodium *What is the pH of a solution that consists of chloride, NH4CI? added?arrow_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_forwardSome kidney stones form by the precipitation of calciumoxalate monohydrate (CaC₂O₄H₂O, Kₛₚ=2.3X10⁻⁹). The pH of urine varies from 5.5 to 7.0, and the average [Ca²⁺] in urine is 2.6X10⁻³ M.(a) If the [oxalic acid] in urine is 3.0X10⁻¹³ M, will kidneystones form at pH =5.5?(b) At pH= 7.0?(c) Vegetarians have a urine pH above 7. Are they more or lesslikely to form kidney stones?arrow_forwardYour answer is partially correct. Ammonia is a convenient buffer system in the slightly basic range. (a) What is the pH of a buffer solution containing 46 g of NH4CI dissolved in 1.00 L of 0.920 M NH3? pH = 9.24 (b) How many moles of acid are required to change the pH of this solution by 0.07 pH units? i 5.5E-2 mol (c) Suppose 4.8 mL of 11.8 M HCI solution is added to 146 mL of the solution of Part (a). Calculate the new pH. pH = i 9.08 eTextbook and Media Save for Later Attempts: 1 of 3 used Submit Answerarrow_forward
- A 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_forwardDetermine the pH during the titration of 71.2 mL of 0.422 M formic acid (Ka = 1.8×10-4) by 0.422 M KOH at the following points. (Assume the titration is done at 25 °C.) (a) Before the addition of any KOH (b) After the addition of 17.0 mL of KOH (c) At the half-equivalence point (the titration midpoint) (d) At the equivalence point (e) After the addition of 107 mL of KOHarrow_forwardA solution NH 3 that contains 78 mL of 0.043 M ammonia, NH 3, is titrated with 0.083 M HCI. The K₁ of ammonia is 1.8×10-5. (a) What volume of 0.083 M HCI would be added to reach the equivalence point? Give the volume in mL. 40.41 mL (b) At the equivalence point, what is the pH of the solution? (Assume that volumes are additive.) 4.56arrow_forward
- General Chemistry - Standalone book (MindTap Cour...ChemistryISBN:9781305580343Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; DarrellPublisher:Cengage LearningChemistry: Principles and ReactionsChemistryISBN:9781305079373Author:William L. Masterton, Cecile N. HurleyPublisher:Cengage Learning