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Chapter 17: Problem 48

In the titration of \(25.00 \mathrm{mL}\) of \(0.100 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}\) calculate the number of milliliters of \(0.200 \mathrm{M} \mathrm{NaOH}\) that must be added to reach a pH of (a) \(3.85,\) (b) 5.25 (c) 11.10.

Short Answer

Expert verified
To reach a pH of 3.85, 5.25, and 11.10, the amount of 0.200 M NaOH that needs to be added will be calculated in the final step based on computations derived from the steps outlined.

Step by step solution

01

Finding Concentration of Hydrogen Ion [H+]

You need to find the concentration of the hydrogen ion [H+] to determine the pH. For a weak acid dissociation is partial and is given by \(CH_3COOH \rightleftharpoons CH_3COO^- + H^+\). The acid dissociation constant \(Ka\) for acetic acid \((CH_3COOH)\) is \(1.8 * 10^{-5}\). Now use the \(Ka\) expression. The \(Ka\) equation is \(Ka=[H^+][CH_3COO^-]/[CH_3COOH]\). Since we are looking for the concentration of \(H^+\) and \(OH^-\) ions before any \(OH^-\) has been added, both \(H^+\) and \(CH_3COO^-\) concentrations are equal and their concentration is \(x\). So, the equation becomes \(Ka=x*x/[CH_3COOH]\)
02

Solving for x

Solving \(Ka=x^2/[CH_3COOH]\) gives \(x^2= Ka*[CH_3COOH]\). Calculating \(x\) gives the \(H^+\) and \(OH^-\) ion concentration before any OH^-\) has been added. Convert \(x\) into pH using the equation \(pH = -log[H^+]\) or \(pH = -log(x)\). After the first few mL of \(NaOH\) have been added, the reaction is still controlled by the presence of the acetic acid, because it is in excess. Now monitor and control pH until its equivalent to the desired pH
03

Finding the volume of NaOH

Calculate the millimoles of \(H^+\) ions using the equation \([H^+]*volume_{acid}\). Then, calculate the millimoles of \(OH^-\) needed to react with \(H^+\) ions to reach the desired pH. The millimoles of \(OH^-\) equals the millimoles of \(H^+\), hence volume of \(OH^-\) equals millimoles of \(OH^-\) divided by the concentration of \(OH^-\). Calculate the volume of \(OH^-\) at pH \(3.85, 5.25, 11.10\) respectively.

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Most popular questions from this chapter

A 20.00 mL sample of \(\mathrm{H}_{3} \mathrm{PO}_{4}(\mathrm{aq})\) requires \(18.67 \mathrm{mL}\) of \(0.1885 \mathrm{M} \mathrm{NaOH}\) for titration from the first to the second equivalence point. What is the molarity of the \(\mathrm{H}_{3} \mathrm{PO}_{4}(\mathrm{aq}) ?\)

The indicator methyl red has a \(\mathrm{pK}_{\mathrm{HIn}}=4.95 . \mathrm{It}\) changes from red to yellow over the pH range from 4.4 to 6.2 (a) If the indicator is placed in a buffer solution of \(\mathrm{pH}=4.55,\) what percent of the indicator will be pre- sent in the acid form, HIn, and what percent will be present in the base or anion form, In \(^{-2}\) (b) Which form of the indicator has the "stronger" (that is, more visible) color- -the acid (red) form or base (yellow) form? Explain.

What stoichiometric concentration of the indicated substance is required to obtain an aqueous solution with the pH value shown: (a) aniline, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\), for \(\mathrm{pH}=8.95 ;(\mathrm{b}) \mathrm{NH}_{4} \mathrm{Cl}\) for \(\mathrm{pH}=5.12 ?\)

Sulfuric acid is a diprotic acid, strong in the first ionization step and weak in the second \(\left(K_{\mathrm{a}_{2}}=1.1 \times 10^{-2}\right)\) By using appropriate calculations, determine whether it is feasible to titrate \(10.00 \mathrm{mL}\) of \(0.100 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) to two distinct equivalence points with \(0.100 \mathrm{M} \mathrm{NaOH}\)

You are asked to prepare a buffer solution with a pH of 3.50. The following solutions, all \(0.100 \mathrm{M},\) are available to you: HCOOH, CH \(_{3} \mathrm{COOH}, \mathrm{H}_{3} \mathrm{PO}_{4}, \mathrm{NaCHOO}\) \(\mathrm{NaCH}_{3} \mathrm{COO},\) and \(\mathrm{NaH}_{2} \mathrm{PO}_{4} . \quad\) Describe how you would prepare this buffer solution. [Hint: What volumes of which solutions would you use?
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