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Chapter 17: Problem 100

A 25.00 -mL sample of \(0.0100 \mathrm{M} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\left(\mathrm{K}_{\mathrm{a}}=\right.\) \(\left.6.3 \times 10^{-5}\right)\) is titrated with \(0.0100 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) Calculate the \(\mathrm{pH}\) (a) of the initial acid solution; (b) after the addition of 6.25 mL of \(0.0100 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) (c) at the equivalence point; (d) after the addition of a total of \(15.00 \mathrm{mL}\) of \(0.0100 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\)

Short Answer

Expert verified
The initial pH of the C6H5COOH solution is calculated to be approximately 2.77. After the addition of 6.25mL of 0.0100M Ba(OH)2, the pH is approximately 4.23. At the equivalence point, the pH is approximately 8.85. After the addition of a total of 15.00 mL of 0.0100 M Ba(OH)2, the pH is approximately 12.74.

Step by step solution

01

- Calculation of initial pH of the acid solution

For a weak acid \(C6H5COOH \rightarrow C6H5COO- + H+ \), the initial \(H+\) ion concentration can be given by \(\sqrt{{ka * [acid]_{initial}}}\), where \(ka = 6.3*10^-5 \) and \([acid]_{initial} = 0.01 M\). Then, using the definition of pH = -log[H+], we get the initial pH value.
02

- Calculation of pH after the addition of 6.25 mL of Ba(OH)2

After adding 6.25 mL of 0.01 M Ba(OH)2, the moles of \(OH-\) ions will neutralize an equivalent quantity of \(H+\) ions and forming \(C6H5COO-\) ion. The remaining \(C6H5COOH\) and \(C6H5COO-\) ion can form a buffer solution. The pH can be calculated by the acid and its conjugate base pair's ratio via the Henderson–Hasselbalch equation:\(pH = pKa + log ( [C6H5COO-]/[C6H5COOH])\).
03

- Calculation of pH at the equivalence point

At the equivalence point, all the weak acid has reacted with the strong base, and only \(C6H5COO-\) ion (conjugate base) remains, which will undergo hydrolysis with water forming \(OH-\) ions and thus forming a basic solution. The \(pOH\) can be calculated by \(\sqrt{{Kw/Ka}} \), where \(Kw = 1x10^-14 \) at 25 degree Celsius. Then, using pH + pOH = 14, we get the pH value.
04

- Calculation of pH after the addition of a total of 15.00ml of 0.0100 M Ba(OH)2

After adding 15ml of 0.01M Ba(OH)2, the strong base is in excess. The \(OH-\) ion concentration can be calculated by the excess moles/volume. Then, pOH = -log[OH-], and using pH + pOH = 14, we get the pH value.

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Most popular questions from this chapter

Even though the carbonic acid-hydrogen carbonate buffer system is crucial to the maintenance of the \(\mathrm{pH}\) of blood, it has no practical use as a laboratory buffer solution. Can you think of a reason(s) for this?

Given \(125 \mathrm{mL}\) of a solution that is \(0.0500 \mathrm{M} \mathrm{CH}_{3} \mathrm{NH}_{2}\) and \(0.0500 \mathrm{M} \mathrm{CH}_{3} \mathrm{NH}_{3}^{+} \mathrm{Cl}^{-}\) (a) Over what pH range will this solution be an effective buffer? (b) What is the buffer capacity of the solution? That is, how many millimoles of strong acid or strong base can be added to the solution before any significant change in pH occurs?

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 \(\mathrm{HNO}_{2}, K_{\mathrm{a}}=7.2 \times 10^{-4}\) \(\mathrm{HNO}_{2}+\mathrm{OH}^{-} \longrightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{NO}_{2}^{-}\)

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 \(0.100 \mathrm{M} \mathrm{NaOH}\)

A very common buffer agent used in the study of biochemical processes is the weak base TRIS, \(\left(\mathrm{HOCH}_{2}\right)_{3} \mathrm{CNH}_{2},\) which has a \(\mathrm{pK}_{\mathrm{b}}\) of 5.91 at \(25^{\circ} \mathrm{C} . \mathrm{A}\) student is given a sample of the hydrochloride of TRIS together with standard solutions of \(10 \mathrm{M}\) NaOH and HCl. (a) Using TRIS, how might the student prepare 1 L of a buffer of \(\mathrm{pH}=7.79 ?\) (b) In one experiment, 30 mmol of protons are released into \(500 \mathrm{mL}\) of the buffer prepared in part (a). Is the capacity of the buffer sufficient? What is the resulting pH? (c) Another student accidentally adds \(20 \mathrm{mL}\) of \(10 \mathrm{M}\) HCl to 500 mL of the buffer solution prepared in part (a). Is the buffer ruined? If so, how could the buffer be regenerated?
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