Concept explainers
(a)
Interpretation:
Molarity of each ion found in
Concept Introduction:
Molarity is quantitatively defined as moles of solute in one liter of solution. For example,
Here,
(a)
Explanation of Solution
Since
Since
Hence, molariry of
(b)
Interpretation:
Molarity of each ion found in
Concept Introduction:
Refer to part (a).
(b)
Explanation of Solution
Since
Since
Hence, molariry of
(c)
Interpretation:
Molarity of each ion found in
Concept Introduction:
Refer to part (a).
(c)
Explanation of Solution
Since
Since
Hence, molariry of
(d)
Interpretation:
Molarity of each ion found in
Concept Introduction:
Refer to part (a).
(d)
Explanation of Solution
Since
Since
Hence, molariry of
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Chapter 15 Solutions
EBK FOUNDATIONS OF COLLEGE CHEMISTRY
- A scientist has synthesized a diprotic organic acid, H2A, with a molar mass of 124.0 g/mol. The acid must be neutralized (forming the potassium salt) for an important experiment. Calculate the volume of 0.221 M KOH that is needed to neutralize 24.93 g of the acid, forming K2A.arrow_forwardIt requires 7.65 mL of 0.100 M HCl solution to neutralize (titrate) 25.00 mL of Ba(OH)2 solution. What is the molar concentration of Ba(OH)2?arrow_forwardIn the following acid-base titration experiment NaOH(aq) + HNO3(aq) ———> NaNO3(aq) + H2O(1). What is the molarity of the HNO3 solution if 18.3 mL of 0.115 M NaOH was used to neutralize 25 mL of HNO3? (a) 0.157 M (b) 0.0842 M (c) 3.978 M (d) 0.0157 Marrow_forward
- How many milliliters of 0.589 M HClO4 are needed to titrate each of the following solutions to the equivalence point?(a) 77.4 mL of 0.942 M KOH mL(b) 36.1 mL of 1.06 M RbOH mL(c) 270.0 mL of a solution that contains 7.08 g of NaOH per literarrow_forwardA chemist is performing a titration in order to determine the amount of sodium hydroxide, NaOH(aq), in 125 mL of an aqueous solution. (a) If 28 mL of a 0.13 M HCl solution was used as a titrant to reach the equivalence point, what was the concentration of sodium hydroxide in the initial solution? M (b) What was the initial pH of the sodium hydroxide solution?arrow_forwardYou wish to determine the concentration of your sulfurous acid. You titrate 27 mL of your acid with 506 mL of 0.23 M aluminum hydroxide. What is the concentration of your sample of sulfurous acid? (A) 0.6 M (B) 0.3 M (C) 0.64 M (D) 0.28 Marrow_forward
- Determine the molar concentration of each ion present in the solutions that result from each of the following mixtures:(Disregard the concentration of H+ and OH- from water and assume that volumes are additive.) (a) 47.9 mL of 0.85 M HCl and 77.6 mL of 1.51 M HCl M H+= M Cl-= (b) 125 mL of 0.63 M CaCl2 and 125 mL of 0.22 M CaCl2 M Ca2+= M Cl-= (c) 31.0 mL of 0.333 M NaOH and 22.7 mL of 0.241 M HCl M Na+= M Cl-= M H+= M OH-= (d) 10.4 mL of 0.602 M H2SO4 and 21.5 mL of 0.151 M NaOH M Na+= M SO42-= M H+= M OH-=arrow_forward(a) A 50.0 mL solution is prepared to be 1.29 M acetylsalicylic acid. In the first step, 8.55 mL of NaOH is titrated into the solution until the pH is exactly 5.0. What is the concentration of the titrant (NaOH)? (b) In the second step, enough 5.85 M nitric acid is added to the solution after the titration in part (a) is complete until the pH is one unit lower than the pKa of acetylsalicylic acid. What volume (mL) of nitric acid was added?arrow_forwardIn the following acid-base titration experiment CH3COOH (aq) + NaOH (aq) ———-> CH3COONa (aq) + H2O (l). What is the molarity of the NaOH solution if 11.6 mL of 3.0 M acetic acid is required to neutralize the sodium hydroxide in 25.00 mL of NaOH solution? (a) 0.65 M (b) 6.5 M (c) 96.6 M (d) 1.392 Marrow_forward
- Determine the molar concentration of each ion present in the solutions that result from each of the following mixtures: (Disregard the concentration of H* and OH* from water and assume that volumes are additive.) (a) 55.6 mL of 0.25 MHCI and 69.8 mL of 1.73 M HCI 0.25 i 1.73 (b) 111 mL of 0.85 M CaCl₂ and 111 mL of 0.21 M CaCl₂ 0.53 i 1.06 i MH* MCI MC₂²+ MCI (c) 37.0 mL of 0.316 MNaOH and 21.9 mL of 0.226 MHCI M Na* MCI MH* MOH (d) 11.9 mL of 0.671 MH₂SO4 and 22.3 mL of 0.146 M NaOH M Na* M50₂²- MH* MOHarrow_forward6.) Calculate the concentration (N) of arsenic acid (H3A504) in a solution if 25mL of that solution required 35.21 mL of 0.1894 M KOH for neutralization?arrow_forwardCake alum (aluminum sulfate) is used as a flocculatingagentin water purification. The “floc” is a gelatinous precipitate of aluminum hydroxide that carries small suspended particles and bacteria out of solution for removal by filtration.(a) The hydroxide is formed by the reaction of the aluminum ion with water. Write the equation for this reaction.(b) The acidity from this reaction is partially neutralized by the sulfate ion. Write an equation for this reaction.arrow_forward
- Chemistry: Principles and PracticeChemistryISBN:9780534420123Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward MercerPublisher:Cengage Learning