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Chapter 15: Problem 88

What quantity (moles) of \(\mathrm{HCl}(g)\) must be added to \(1.0 \mathrm{~L}\) of 2.0 \(M \mathrm{NaOH}\) to achieve a pH of \(0.00 ?\) (Neglect any volume changes.)

Short Answer

Expert verified
To achieve a pH of \(0.00\) in \(1.0 \mathrm{~L}\) of 2.0 \(M \mathrm{NaOH}\), you need to add \(3.0 \mathrm{~moles}\) of \(\mathrm{HCl}(g)\).

Step by step solution

01

Calculate initial moles of \(\mathrm{NaOH}\)

The initial moles of \(\mathrm{NaOH}\) can be calculated using the initial volume and molarity. Initial moles of \(\mathrm{NaOH}\) = Molarity × Volume Initial moles of \(\mathrm{NaOH}\) = \(2.0M \times 1.0L = 2.0\) moles
02

Calculate the final concentration of hydrogen ions \(([H^+])\)

Since we want a pH of \(0.00\), we will find out the final concentration of hydrogen ions using the pH formula: \(pH = -\log[H^+]\) We need to rearrange the formula to solve for \([H^+]\): \([H^+] = 10^{-pH}\) Hence, \([H^+] = 10^{-0} = 1 \mathrm{M}\)
03

Find moles of \(H^+\) ions required to achieve the desired pH

Now we will find the moles of \(H^+\) ions required to achieve the desired pH using the final concentration of \(H^+\) ions and the final volume of the solution: Moles of \(H^+\) ions = Concentration × Volume Moles of \(H^+\) ions = \(1 \mathrm{M} \times 1.0 \mathrm{L} = 1.0\) moles
04

Determine moles of \(HCl\) required to neutralize the given \(\mathrm{NaOH}\)

In order to neutralize the \(2.0\) moles of \(\mathrm{NaOH}\), we need an equal number of moles of \(\mathrm{HCl}\) (both being strong acid and strong base; they neutralize each other): Neutralization reaction: \(\mathrm{NaOH + HCl \rightarrow NaCl + H_2O}\) Moles of \(\mathrm{HCl}\) required for neutralization = Moles of \(\mathrm{NaOH}\) Moles of \(\mathrm{HCl}\) required for neutralization = \(2.0\) moles
05

Calculate the total moles of \(\mathrm{HCl}\) to be added

Now, we need to add the moles of \(\mathrm{HCl}\) required for neutralization and the moles of \(H^+\) ions required to achieve the desired pH: Total moles of \(\mathrm{HCl}\) = Moles of \(\mathrm{HCl}\) required for neutralization + Moles of \(H^+\) ions required for desired pH Total moles of \(\mathrm{HCl}\) = \(2.0 \mathrm{~moles} + 1.0 \mathrm{~moles} = 3.0 \mathrm{~moles}\) Therefore, \(3.0 \mathrm{~moles}\) of \(\mathrm{HCl}(g)\) must be added to \(1.0 \mathrm{~L}\) of 2.0 \(M \mathrm{NaOH}\) to achieve a pH of \(0.00\).

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Most popular questions from this chapter

Consider the titration of \(80.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) by \(0.400 M\) HCl. Calculate the \(\mathrm{pH}\) of the resulting solution after the following volumes of \(\mathrm{HCl}\) have been added. a. \(0.0 \mathrm{~mL}\) b. \(20.0 \mathrm{~mL}\) c. \(30.0 \mathrm{~mL}\) d. \(40.0 \mathrm{~mL}\) e. \(80.0 \mathrm{~mL}\)

Calculate the \(\mathrm{pH}\) at the halfway point and at the equivalence point for each of the following titrations. a. \(100.0 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{HC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}\left(K_{\mathrm{a}}=6.4 \times 10^{-5}\right)\) titrated by \(0.10 \mathrm{M} \mathrm{NaOH}\) b. \(100.0 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\left(K_{\mathrm{b}}=5.6 \times 10^{-4}\right)\) titrated by \(0.20 \mathrm{M} \mathrm{HNO}_{3}\) c. \(100.0 \mathrm{~mL}\) of \(0.50 \mathrm{M} \mathrm{HCl}\) titrated by \(0.25 \mathrm{M} \mathrm{NaOH}\)

Consider the titration of \(40.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{HClO}_{4}\) by \(0.100 M \mathrm{KOH}\). Calculate the \(\mathrm{pH}\) of the resulting solution after the following volumes of KOH have been added. a. \(0.0 \mathrm{~mL}\) b. \(10.0 \mathrm{~mL}\) c. \(40.0 \mathrm{~mL}\) d. \(80.0 \mathrm{~mL}\) e. \(100.0 \mathrm{~mL}\)

Repeat the procedure in Exercise 61 , but for the titration of \(25.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{NH}_{3}\left(K_{\mathrm{b}}=1.8 \times 10^{-5}\right)\) with \(0.100 \mathrm{M}\) \(\mathrm{HCl}\)

Calculate the ratio \(\left[\mathrm{NH}_{3}\right] /\left[\mathrm{NH}_{4}{ }^{+}\right]\) in ammonia/ammonium chloride buffered solutions with the following \(\mathrm{pH}\) values: a. \(\mathrm{pH}=9.00\) b. \(\mathrm{pH}=8.80\) c. \(\mathrm{pH}=10.00\) d. \(\mathrm{pH}=9.60\)
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