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Chapter 17: Problem 71

You are asked to bring the \(\mathrm{pH}\) of \(0.500 \mathrm{L}\) of \(0.500 \mathrm{M}\) \(\mathrm{NH}_{4} \mathrm{Cl}(\mathrm{aq})\) to 7.00 How many drops \((1 \text { drop }=0.05 \mathrm{mL})\) of which of the following solutions would you use: \(10.0 \mathrm{M} \mathrm{HCl}\) or \(10.0 \mathrm{M} \mathrm{NH}_{3} ?\)

Short Answer

Expert verified
Approximately 496 drops of \(10.0 M NH_3\) solution are needed to bring the pH of the \(0.500 L\) of \(0.500 M NH_4Cl\) solution to 7.00.

Step by step solution

01

Compute the initial moles of \(NH_4Cl\)

First, find the number of moles of \(NH_4Cl\) present in the 0.500 L solution by multiplying its volume by its molarity: \(0.500L \times 0.500mol/L = 0.250mol\) of \(NH_4Cl\).
02

Calculate the \(Kb\) of \(NH_3\)

Next, using the \(Kw\) (\(1.00 \times 10^{-14}\)) and \(Ka\) (\(5.56 \times 10^{-10}\)) values for \(NH_4^+\), calculate the \(Kb\) for \(NH_3\) with the formula \[ Kb = Kw / Ka \] This gives \(Kb = (1.00 \times 10^{-14}) / (5.56 \times 10^{-10}) ≈ 1.80 \times 10^{-5}\].
03

Create an ICE table

The chemical equation for the dissociation of \(NH_4Cl\) is \[ NH_4^+ + H2O ⇌ NH3 + H3O^+ \] The ICE table would look like this: \nI 0.250 - 0 - \nC -x - +x +x \nE 0.250-x - x x \nKb = [ \(NH_3\) ][ \(H_3O^+\) ] / [ \(NH_4^+\) ] \nplug in values and solve for x \n\(1.80 \times 10^{-5}\) = (x)(x) / (0.250 - x)\n\nAssuming x << 0.250, the equation simplifies to \[ x = √[(0.250) (1.80 \times 10^{-5})] = 2.12 \times 10^{-3} M \] The concentration of \(NH_3\) needed to bring pH to 7.00 is \(2.12 \times 10^{-3} M\).
04

Determine which solution is needed

Since the \(NH_3\) concentration is lower than \(NH_4Cl\) concentration, a base needs to be added. Thus, drops of \(10.0 M NH_3\) solution should be added.
05

Calculate the number of drops

Finally, calculate the number of moles of \(NH_3\) needed, which is equal to the moles of \(NH_4Cl\), using the formula: \[ molesNH_3 = molesNH_{4}Cl - [NH_3] = 0.250mol - 2.12 \times 10^{-3} mol ≈ 0.248mol \] To find the volume of \(10.0 M NH_3\) needed, divide the moles of \(NH_3\) by its molarity, which gives 0.0248 L or 24.8 mL. Now, to get the number of drops, divide 24.8 mL by the volume per drop (0.05 mL/drop), which equals to about 496 drops.

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Most popular questions from this chapter

Lactic acid, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH},\) is found in sour milk. A solution containing \(1.00 \mathrm{g} \mathrm{NaCH}_{3} \mathrm{CH}_{2} \mathrm{COO}\) in 100.0 \(\mathrm{mL}\) of \(0.0500 \mathrm{M} \mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{3}\) has a \(\mathrm{pH}=4.11 .\) What is \(K_{\mathrm{a}}\) of lactic acid?

Rather than calculate the \(\mathrm{pH}\) for different volumes of titrant, a titration curve can be established by calculating the volume of titrant required to reach certain \(\mathrm{pH}\) values. Determine the volumes of \(0.100 \mathrm{M} \mathrm{NaOH}\) required to reach the following pH values in the titration of \(20.00 \mathrm{mL}\) of \(0.150 \mathrm{M} \mathrm{HCl}: \mathrm{pH}=\) (a) 2.00 (b) \(3.50 ;\) (c) \(5.00 ;\) (d) \(10.50 ;\) (e) \(12.00 .\) Then plot the titration curve.

Piperazine is a diprotic weak base used as a corrosion inhibitor and an insecticide. Its ionization is described by the following equations. \(\mathrm{HN}\left(\mathrm{C}_{4} \mathrm{H}_{8}\right) \mathrm{NH}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons\) \(\left[\mathrm{HN}\left(\mathrm{C}_{4} \mathrm{H}_{8}\right) \mathrm{NH}_{2}\right]^{+}+\mathrm{OH}^{-} \quad \mathrm{p} K_{\mathrm{b}_{1}}=4.22\) \(\left[\mathrm{HN}\left(\mathrm{C}_{4} \mathrm{H}_{8}\right) \mathrm{NH}_{2}\right]^{+}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons\) \(\left[\mathrm{H}_{2} \mathrm{N}\left(\mathrm{C}_{4} \mathrm{H}_{8}\right) \mathrm{NH}_{2}\right]^{2+}+\mathrm{OH}^{-} \quad \mathrm{p} K_{\mathrm{b}_{2}}=8.67\) . The piperazine used commercially is a hexahydrate, \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{N}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O} .\) A \(1.00-\mathrm{g}\) sample of this hexahydrate is dissolved in \(100.0 \mathrm{mL}\) of water and titrated with 0.500 M HCl. Sketch a titration curve for this titration, indicating (a) the initial \(\mathrm{pH} ;\) (b) the pH at the halfneutralization point of the first neutralization; (c) the volume of \(\mathrm{HCl}(\text { aq })\) required to reach the first equivalence point; (d) the pH at the first equivalence point; (e) the \(\mathrm{pH}\) at the point at which the second step of the neutralization is half-completed; (f) the volume of \(0.500 \mathrm{M} \mathrm{HCl}(\) aq) required to reach the second equivalence point of the titration; (g) the pH at the second equivalence point.

Calculate the \(\mathrm{pH}\) at the points in the titration of \(25.00 \mathrm{mL}\) of \(0.132 \mathrm{M} \mathrm{HNO}_{2}\) when (a) \(10.00 \mathrm{mL}\) and (b) 20.00 mL of 0.116 M \(\mathrm{HNO}_{2}\) when \((\mathrm{a}) 10.00 \mathrm{mL}\) and (b) \(20.00 \mathrm{mL}\) of \(0.116 \mathrm{M}\) NaOH have been added. For \(\mathrm{HNO}_{2}, K_{\mathrm{a}}=7.2 \times 10^{-4}\) \(\mathrm{HNO}_{2}+\mathrm{OH}^{-} \longrightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{NO}_{2}^{-}\)

Explain whether the equivalence point of each of the following titrations should be below, above, or at pH 7: (a) \(\mathrm{NaHCO}_{3}(\text { aq) titrated with } \mathrm{NaOH}(\mathrm{aq}) ; \text { (b) } \mathrm{HCl}(\mathrm{aq})\) titrated with \(\mathrm{NH}_{3}(\mathrm{aq}) ;\) (c) KOH(aq) titrated with HI(aq).
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