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Chapter 4: Problem 70

What volume of each of the following bases will react completely with \(25.00 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{HCl}\) ? a. \(0.100 \mathrm{M} \mathrm{NaOH}\) b. \(0.0500 \mathrm{M} \mathrm{Sr}(\mathrm{OH})_{2}\) c. \(0.250 \mathrm{M} \mathrm{KOH}\)

Short Answer

Expert verified
The volumes required to completely react with \(25.00 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{HCl}\) are: a) \(50.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{NaOH}\) b) \(50.0 \mathrm{~mL}\) of \(0.0500 \mathrm{M} \mathrm{Sr}(\mathrm{OH})_{2}\) c) \(20.0 \mathrm{~mL}\) of \(0.250 \mathrm{M} \mathrm{KOH}\)

Step by step solution

01

Write balanced chemical equations

For each base, write the balanced chemical equation with HCl. a) NaOH + HCl -> NaCl + H2O b) Sr(OH)2 + 2HCl -> SrCl2 + 2H2O c) KOH + HCl -> KCl + H2O
02

Calculate moles of HCl

The initial volume of the given HCl solution is 25.00 mL, and its concentration is 0.200 M. The number of moles of HCl can be calculated as follows: moles of HCl = volume (L) × concentration (mol/L) moles of HCl = \(0.02500 L × 0.200 mol L^{-1} = 0.00500 mol\)
03

Calculate moles of each base

Using the mole ratios from the balanced chemical equations from Step 1, calculate the moles of each base needed for the reaction. a) NaOH: 1 mol NaOH reacts with 1 mol HCl, so moles of NaOH = moles of HCl = 0.00500 mol b) Sr(OH)2: 1 mol Sr(OH)2 reacts with 2 mol HCl, so moles of Sr(OH)2 = (0.00500 mol HCl) / 2 = 0.00250 mol c) KOH: 1 mol KOH reacts with 1 mol HCl, so moles of KOH = moles of HCl = 0.00500 mol
04

Calculate volume of each base

Finally, use the given concentration for each base to find the required volume that will completely react with the given HCl solution. a) Volume of NaOH: \(v = \dfrac{0.00500 mol}{0.100 mol L^{-1}} = 0.0500 L = 50.0 mL\) b) Volume of Sr(OH)2: \(v = \dfrac{0.00250 mol}{0.0500 mol L^{-1}} = 0.0500 L = 50.0 mL\) c) Volume of KOH: \(v = \dfrac{0.00500 mol}{0.250 mol L^{-1}} = 0.0200 L = 20.0 mL\) To summarize, the volumes required to completely react with 25.00 mL of 0.200 M HCl are: a) 50.0 mL of 0.100 M NaOH b) 50.0 mL of 0.0500 M Sr(OH)2 c) 20.0 mL of 0.250 M KOH

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Most popular questions from this chapter

A student mixes four reagents together, thinking that the solutions will neutralize each other. The solutions mixed together are \(50.0 \mathrm{~mL}\) of \(0.100 \mathrm{M}\) hydrochloric acid, \(100.0 \mathrm{~mL}\) of \(0.200 \mathrm{M}\) of nitric acid, \(500.0 \mathrm{~mL}\) of \(0.0100 \mathrm{M}\) calcium hydroxide, and \(200.0 \mathrm{~mL}\) of \(0.100 M\) rubidium hydroxide. Did the acids and bases exactly neutralize each other? If not, calculate the concentration of excess \(\mathrm{H}^{+}\) or \(\mathrm{OH}^{-}\) ions left in solution.

A \(100.0\) -mLaliquot of \(0.200 M\) aqueous potassium hydroxide is mixed with \(100.0 \mathrm{~mL}\) of \(0.200 \mathrm{M}\) aqueous magnesium nitrate. a. Write a balanced chemical equation for any reaction that occurs. b. What precipitate forms? c. What mass of precipitate is produced? d. Calculate the concentration of each ion remaining in solution after precipitation is complete.

Specify which of the following are oxidation-reduction reactions, and identify the oxidizing agent, the reducing agent, the substance being oxidized, and the substance being reduced. a. \(\mathrm{Cu}(s)+2 \mathrm{Ag}^{+}(a q) \rightarrow 2 \mathrm{Ag}(s)+\mathrm{Cu}^{2+}(a q)\) b. \(\mathrm{HCl}(g)+\mathrm{NH}_{3}(\mathrm{~g}) \rightarrow \mathrm{NH}_{4} \mathrm{Cl}(s)\) c. \(\mathrm{SiCl}_{4}(l)+2 \mathrm{H}_{2} \mathrm{O}(l) \rightarrow 4 \mathrm{HCl}(a q)+\mathrm{SiO}_{2}(s)\) d. \(\mathrm{SiCl}_{4}(l)+2 \mathrm{Mg}(s) \rightarrow 2 \mathrm{MgCl}_{2}(s)+\mathrm{Si}(s)\) e. \(\mathrm{Al}(\mathrm{OH})_{4}^{-}(a q) \rightarrow \mathrm{AlO}_{2}^{-}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)\)

When hydrochloric acid reacts with magnesium metal, hydrogen gas and aqueous magnesium chloride are produced. What volume of \(5.0 \mathrm{M} \mathrm{HCl}\) is required to react completely with \(3.00 \mathrm{~g}\) of magnesium?

Complete and balance each acid-base reaction. a. \(\mathrm{H}_{3} \mathrm{PO}_{4}(a q)+\mathrm{NaOH}(a q) \rightarrow\) Contains three acidic hydrogens b. \(\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{Al}(\mathrm{OH})_{3}(s) \rightarrow\) Contains two acidic hydrogens c. \(\mathrm{H}_{2} \mathrm{Se}(a q)+\mathrm{Ba}(\mathrm{OH})_{2}(a q) \rightarrow\) Contains two acidic hydrogens d. \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(a q)+\mathrm{NaOH}(a q) \rightarrow\) Contains two acidic hydrogens
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