What volume of water must be added to 500 mL of HCl having pH 150 in order to

What volume of water must be added to 500 ml of hcl

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8. What volume of water must be added to 50.0 mL of HCl having pH 1.50, in order to produce a solution of HCl having pH 2.50? A. 100 mL B. 950 mL C. 25 mL D. 450 mL E. 250 mL Convert pH or pOH into [H+] or [OH-], respectively, before doing anycalculations. One cannot work with pH or pOH. Initial: [H+] = 10-pH= 10-1.50= 0.03162M Final: [H+] = 10-pH= 10-2.50= 0.003162M For dilution, one may use M1V1= M2V2. For other calculations, go through moles. M1V1= M2V20.03162M x 0.050L = 0.003162M x XL X = 0.50L = 500 mL total volume at the end. But there are 50 mL already present. Hence 450 mL of water needs to be added.
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26 Chem 162-2011 Hourly Exam II + Answers Chapter 15B - Applic. Of Acid & Base Equilibria (Buffers & Titrations) Acid and base equilibria calculations 19mod. 5.00 mL of a solution of NaOH having pH 13.00 is added to 495 mL of water at 25oC. What is the pH of the resultant solution? X
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Chem 162-2015 Chapter 16 Tro Non-Acid-Base Equilibria 27 CHEM 162-2009 HOURLY EXAM II + ANSWERS CHAPTER 15A - ACID & BASE EQUILIBRIA ACID AND BASE EQUILIBRIA CALCULATIONS 19. At 50 o C, the pH of pure water is 6.63. Calculate K w for water at 50 o C. A. 1.1 x 10 -14 B. 2.2 x 10 -14 C . 5.5 x 10 -14 D. 3.3 x 10 -14 E. 4.4 x 10 -14 + H 2 O + H 2 O H 3 O + + OH - Water autoionization is an endothermic reaction. At 25 o C K w = 1 x 10 -14 Therefore, at 50 o C K w > 1 x 10 -14 . pH = 6.63 [H + ] = 10 -6.63 = 2.344 x 10 -7 [H 2 O] + [H 2 O] [H 3 O + ] + [OH - ] Note that in pure H 2 O there is an equal concentration of H 3 O + & OH - . [H 3 O + ] = [OH - ] [H 3 O + ][OH - ] = K w [2.344 x 10 -7 ][ 2.344 x 10 -7 ] = 5.50 x 10 -14 = K w
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Chem 162-2015 Chapter 16 Tro Non-Acid-Base Equilibria 28 TABLE OF CONTENTS FOR ACID-BASE EQUILIBRIA PROBLEMS p. 30. Given initial acid concentration and K a , find equilibrium concentrations. p. 31. Given initial acid concentration and K a , find equilibrium concentrations (percent dissociation). p. 33. Given initial acid concentration and equilibrium concentration (% dissociation), find K a . p. 34. Given initial acid concentration and equilibrium concentration (pH), find K a . p. 35. Given initial base concentration and equilibriumconcentration (pH), find K b . p. 36. Given initial base concentration and K b , find equilibrium concentration (pH). p. 38. Given initial polyprotic acid concentration and 3 Ka’s, find all equilibrium concentrations. p. 39. Given initial diprotic acid concentration and 2 Ka’s, find all equilibrium concentrations. p. 43. Given initial basic salt concentration and Ka, find equilibrium concentration (pH). p. 44. Given initial basic salt concentration and equilibrium concentration (pH), find K a . p. 45. Given initial acidic salt concentration and K b , find equilibrium concentration (pH). p. 46. Given equilibrium acidic salt concentration (pH) and K b , find initial concentration.
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Chem 162-2015 Chapter 16 Tro Non-Acid-Base Equilibria 29 REACTIONS OF ACIDS AND BASES WITH WATER A or CA or B or CB + H 2 O 2 Products ACID + WATER H H — A + :A - + CONJUGATE ACID + WATER H H H R—N—H + R—N: H H BASE + WATER H ..
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