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Chapter 15: Problem 23

Determine the molar concentration of each ion present in the solutions that result from each of the following mixtures: (Disregard the concentration of \(\mathrm{H}^{+}\)and \(\mathrm{OH}^{-}\)from water and assume that volumes are additive.) (a) \(55.5 \mathrm{~mL}\) of \(0.50 \mathrm{M} \mathrm{HCl}\) and \(75.0 \mathrm{~mL}\) of \(1.25 \mathrm{M} \mathrm{HCl}\) (b) \(125 \mathrm{~mL}\) of \(0.75 \mathrm{M} \mathrm{CaCl}_{2}\) and \(125 \mathrm{~mL}\) of \(0.25 \mathrm{MCaCl}_{2}\) (c) \(35.0 \mathrm{~mL}\) of \(0.333 \mathrm{M} \mathrm{NaOH}\) and \(22.5 \mathrm{~mL}\) of \(0.250 M \mathrm{HCl}\) (d) \(12.5 \mathrm{~mL}\) of \(0.500 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) and \(23.5 \mathrm{~mL}\) of \(0.175 M \mathrm{NaOH}\)

Short Answer

Expert verified
(a) [H+] = [Cl-] = 0.93 M(b) [Ca2+] = 0.5 M, [Cl-] = 1.0 M(c) [NaOH] = 0.105 M(d) [H2SO4] = 0.1165 M

Step by step solution

01

Calculate total volume for each mixture

(a) Total volume = 55.5 mL + 75.0 mL = 130.5 mL (b) Total volume = 125 mL + 125 mL = 250 mL (c) Total volume = 35.0 mL + 22.5 mL = 57.5 mL (d) Total volume = 12.5 mL + 23.5 mL = 36.0 mL
02

Calculate moles of solutes

(a) Moles of HCl = (0.50 M * 55.5 mL/1000) + (1.25 M * 75.0 mL/1000) (b) Moles of CaCl2 = (0.75 M * 125 mL/1000) + (0.25 M * 125 mL/1000) (c) Moles of NaOH = 0.333 M * 35.0 mL/1000, Moles of HCl = 0.25 M * 22.5 mL/1000 (d) Moles of H2SO4 = 0.50 M * 12.5 mL/1000, Moles of NaOH = 0.175 M * 23.5 mL/1000
03

Calculate molar concentrations in final solutions

(a) Total moles of HCl = (0.02775 + 0.09375) = 0.1215 moles. Molar concentration of HCl = 0.1215 moles / 0.1305 L = 0.93 M. Since HCl dissociates to H+ and Cl-, [H+] = [Cl-] = 0.93 M (b) Total moles of CaCl2 = (0.09375 + 0.03125) = 0.125 moles. Molar concentration of CaCl2 = 0.125 moles / 0.25 L = 0.5 M. Since CaCl2 dissociates into Ca2+ and 2Cl-, [Ca2+] = 0.5 M and [Cl-] = 1.0 M (c) Moles of NaOH = 0.011655 moles, Moles of HCl = 0.005625 moles. Net moles of NaOH (excess) = 0.011655 - 0.005625 = 0.00603 moles. Molar concentration of NaOH in final solution = 0.00603 moles / 0.0575 L = 0.105 M (d) Moles of H2SO4 = 0.00625 moles, Moles of NaOH = 0.0041125 moles. Net moles of H2SO4 (excess) = 0.00625 - 0.0041125/2 = 0.004194 moles. Molar concentration of H2SO4 in final solution = 0.004194 moles / 0.036 L = 0.1165 M.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

solution chemistry
Solution chemistry involves the study of substances dissolved in a solvent, usually water. When compounds like HCl, CaCl2, NaOH, and H2SO4 dissolve, they dissociate into their constituent ions. It's critical to understand how these solutes behave in solution:

  • Ionic compounds like NaCl will break into Na+ and Cl- ions.
  • Covalent compounds like HCl will dissociate into H+ and Cl- ions when dissolved in water.
In this exercise, calculating the molar concentration relies on knowing how these substances dissociate. We determine the total volume after mixing solutions and then compute the total amount of moles for each ion. This gives a clear picture of the resulting ion concentrations in the solution.
molarity
Molarity (M) is a measure of concentration, defined as moles of solute per liter of solution. Here is the calculation step-by-step:

  • Convert volumes to liters.
  • Multiply the molarity of each solution by its volume to find the moles of solute.
For example, in exercise (a), we mix 55.5 mL of 0.50 M HCl and 75.0 mL of 1.25 M HCl:

  • The number of moles from 0.50 M HCl: \(0.50 \frac{moles}{L} \times \frac{55.5}{1000} L = 0.02775 \, moles\)
  • The number of moles from 1.25 M HCl: \(1.25 \frac{moles}{L} \times \frac{75.0}{1000} L = 0.09375 \, moles\)
Add these moles together, then divide by the total volume in liters to find the final molarity: \( \frac{0.02775 + 0.09375}{0.1305} = 0.93 \, M\)This calculation helps determine how concentrated specific ions are within the resulting solution.
acid-base reaction
Acid-base reactions are vital in chemistry as they involve the transfer of protons (H+). Acids like HCl donate H+ ions, while bases like NaOH accept them. When they react, they usually form water and a salt:

  • HCl (acid) dissociates to H+ and Cl-.
  • NaOH (base) dissociates to Na+ and OH-.
If mixed, the H+ from HCl neutralizes the OH- from NaOH, forming water. Any remaining (excess) ions contribute to the solution's molarity:

  • For instance, in part (c) NaOH is in excess:
  • Mixing calculations: \(0.011655 - 0.005625 = 0.00603 \, moles \, of \, NaOH \, left\)
The leftover NaOH defines the final NaOH molarity: \( \frac{0.00603}{0.0575} \, L = 0.105 \, M\)
This understanding is crucial for predicting the behavior of mixed acid-base solutions.
stoichiometry
Stoichiometry involves using balanced chemical equations to determine the relationships between reactants and products. In solution chemistry, it helps predict the amounts of substances involved:

  • Calculate moles from given molarity and volume.
  • Determine the reaction's limiting reactant and excess reactants.
For example, in problem (d), we have H2SO4 reacting with NaOH:

  • H2SO4 + 2NaOH \(\rightarrow\) Na2SO4 + 2H2O
  • Moles of H2SO4: \(0.50 \frac{moles}{L} \times 0.0125 \, L = 0.00625 \, moles\)
  • Moles of NaOH: \(0.175 \frac{moles}{L} \times 0.0235 \, L = 0.0041125 \, moles\)
  • Reacting ratio check shows H2SO4 is in excess.
  • Calculate remaining moles and use total volume to find concentration.
This process shows how reactant amounts dictate the resulting solution composition.

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Most popular questions from this chapter

One of the tests used to detect the presence of LSD or some of the other psilocybin-like alkaloids often found in psychedelic mushrooms is the Ehrlich test. For this test a solution of \(p\) dimethylaminobenzaldehyde, \(\mathrm{C}_{9} \mathrm{H}_{11} \mathrm{NO}\), in \(3.25 \mathrm{M} \mathrm{HCI}\) is used to detect the presence of any of these compounds. How many \(\mathrm{mL}\) of concentrated \(\mathrm{HCI}(12.1 M)\) are required to make \(3.00 \mathrm{~L}\) of \(3.25 \mathrm{MHCI}\) for preparing Ehrlich reagent?

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What volume of \(\mathrm{H}_{2}\) gas, measured at \(27^{\circ} \mathrm{C}\) and 700 . torr, can be obtained by reacting \(5.00 \mathrm{~g}\) of zinc metal with \(100 . \mathrm{mL}\) of \(0.350\) \(M \mathrm{HCl}\) ? The equation is $$ \mathrm{Zn}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g) $$

Determine whether each of the following is a strong acid, weak acid, strong base, or weak base. Then write an equation describing the process that occurs when the substance is dissolved in water. (a) \(\mathrm{NH}_{3}\) (b) \(\mathrm{HCl}\) (c) \(\mathrm{KOH}\) (d) \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\)
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