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Chapter 4: Problem 90

Calculate the concentration (in molarity) of a \(\mathrm{NaOH}\) solution if \(25.0 \mathrm{~mL}\) of the solution are needed to neutralize \(17.4 \mathrm{~mL}\) of a \(0.312 \mathrm{M} \mathrm{HCl}\) solution.

Short Answer

Expert verified
The molarity of the NaOH solution is \(0.217152 M\)

Step by step solution

01

Convert volumes from mL to L

First convert the given volumes from milliliters to liters because molarity is usually expressed in terms of liters. This can be done by dividing the given volume by 1000, as 1L equals 1000mL. So, \(25.0 \mathrm{~mL}\) becomes \(0.025 \mathrm{~L}\) and \(17.4\mathrm{~mL}\) becomes \(0.0174\mathrm{~L}\)
02

Understand the chemical reaction

In this neutralization reaction, Sodium Hydroxide (NaOH) reacts with Hydrochloric Acid (HCl) to produce water and Sodium Chloride (NaCl), given by the balanced equation: \(NaOH + HCl → H2O + NaCl\). This reaction is a 1:1 reaction, so the number of moles for HCl equals that of NaOH.
03

Calculate the number of moles of HCl

The number of moles of HCl can be calculated using the given molarity and volume of HCl. The molarity of a solution equals the number of moles divided by the volume. So, number of moles equals molarity multiplied by volume, which gives \(0.312 M \times 0.0174 L = 0.0054288 moles\) of HCl.
04

Find the molarity of NaOH

Because the reaction is 1:1, the number of moles of HCl equals the number of moles of NaOH. So, there are \(0.0054288 \mathrm{moles}\) of NaOH. The molarity of NaOH can then be calculated by dividing the number of moles by the volume, which gives \( \frac{0.0054288 moles}{0.025 L} = 0.217152 M\)

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Most popular questions from this chapter

Explain how you would prepare potassium iodide (KI) by means of (a) an acid- base reaction and (b) a reaction between an acid and a carbonate compound.

Write the equation that enables us to calculate the concentration of a diluted solution. Give units for all the terms.

Which of the following metals can react with water? (a) \(\mathrm{Au},(\mathrm{b}) \mathrm{Li},(\mathrm{c}) \mathrm{Hg},(\mathrm{d}) \mathrm{Ca},(\mathrm{e}) \mathrm{Pt}\)

Classify the following reactions according to the types discussed in the chapter. (a) \(\mathrm{Cl}_{2}+2 \mathrm{OH}^{-} \longrightarrow \mathrm{Cl}^{-}+\mathrm{ClO}^{-}+\mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{Ca}^{2+}+\mathrm{CO}_{3}^{2-} \longrightarrow \mathrm{CaCO}_{3}\) (c) \(\mathrm{NH}_{3}+\mathrm{H}^{+} \longrightarrow \mathrm{NH}_{4}^{+}\) (d) \(2 \mathrm{CCl}_{4}+\mathrm{CrO}_{4}^{2-} \longrightarrow\) \(2 \mathrm{COCl}_{2}+\mathrm{CrO}_{2} \mathrm{Cl}_{2}+2 \mathrm{Cl}^{-}\) (e) \(\mathrm{Ca}+\mathrm{F}_{2} \longrightarrow \mathrm{CaF}_{2}\) (f) \(2 \mathrm{Li}+\mathrm{H}_{2} \longrightarrow 2 \mathrm{LiH}\) (g) \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}+\mathrm{Na}_{2} \mathrm{SO}_{4} \longrightarrow 2 \mathrm{NaNO}_{3}+\mathrm{BaSO}_{4}\) (h) \(\mathrm{CuO}+\mathrm{H}_{2} \longrightarrow \mathrm{Cu}+\mathrm{H}_{2} \mathrm{O}\) (i) \(\mathrm{Zn}+2 \mathrm{HCl} \longrightarrow \mathrm{ZnCl}_{2}+\mathrm{H}_{2}\) (j) \(2 \mathrm{FeCl}_{2}+\mathrm{Cl}_{2} \longrightarrow 2 \mathrm{FeCl}_{3}\) (k) \(\mathrm{LiOH}+\mathrm{HNO}_{3} \longrightarrow \mathrm{LiNO}_{3}+\mathrm{H}_{2} \mathrm{O}\)

Chemical tests of four metals \(\mathrm{A}, \mathrm{B}, \mathrm{C},\) and \(\mathrm{D}\) show the following results. (a) Only \(\mathrm{B}\) and \(\mathrm{C}\) react with \(0.5 \mathrm{M} \mathrm{HCl}\) to give \(\mathrm{H}_{2}\) gas. (b) When \(\mathrm{B}\) is added to a solution containing the ions of the other metals, metallic \(\mathrm{A}, \mathrm{C},\) and \(\mathrm{D}\) are formed. (c) A reacts with \(6 M \mathrm{HNO}_{3}\) but \(\mathrm{D}\) does not. Arrange the metals in the increasing order as reducing agents. Suggest four metals that fit these descriptions.
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