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Chapter 15: Problem 111

What volume of \(0.0100 M\) NaOH must be added to \(1.00 \mathrm{~L}\) of \(0.0500 \mathrm{M} \mathrm{HOCl}\) to achieve a \(\mathrm{pH}\) of \(8.00 ?\)

Short Answer

Expert verified
To achieve a pH of \(8.00\), add \(1.00 \times 10^{-4}~L\) of \(0.0100~M\) NaOH to \(1.00~L\) of \(0.0500~M~HOCl\).

Step by step solution

01

Calculate the pOH and the concentration of \(OH^-\) ions required

Since we are given the target pH, we can calculate the pOH and the concentration of hydroxide ions, \(OH^-\), needed to achieve this pH. $$pOH = 14 - pH = 14 - 8.00 = 6.00$$ Now, using the pOH, we can find the concentration of hydroxide ions present: $$[OH^-] = 10^{-pOH} = 10^{-6.00} = 1.00 \times 10^{-6}~M$$
02

Calculate the moles of \(OH^-\) ions required

Now we have the target \(OH^-\) concentration needed to achieve a pH of 8.00, and since the volume of the \(HOCl\) solution is given, we can compute the moles of \(OH^-\) ions required: $$ moles~of~OH^- = (1.00 \times 10^{-6}~M) \times (1.00~L) = 1.00 \times 10^{-6}~moles$$
03

Calculate the volume of \(0.0100 M\) NaOH required

Since we know the number of moles of \(OH^-\) ions required and the concentration of NaOH, we can use the following equation to calculate the volume of NaOH required: $$Volume~of~NaOH = \frac{moles~of~OH^-}{[OH^-]~in~NaOH}$$ Plugging in the values, we get: $$Volume~of~NaOH = \frac{1.00 \times 10^{-6}~moles}{0.0100 M} = 1.00 \times 10^{-4}~L$$ So, to achieve a pH of 8.00, \(1.00 \times 10^{-4}~L\) of \(0.0100~M\) NaOH should be added to \(1.00~L\) of \(0.0500~M~HOCl\).

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Most popular questions from this chapter

Consider the following two acids: $$ \mathrm{p} K_{\mathrm{a}_{1}}=2.98 ; \mathrm{p} K_{\mathrm{a}_{2}}=13.40 $$ $$ \begin{array}{l} \mathrm{HO}_{2} \mathrm{CCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H} \\ \text { Adipic acid } \quad \mathrm{p} K_{\mathrm{a}_{1}}=4.41 ; \mathrm{p} K_{\mathrm{a}_{2}}=5.28 \end{array} $$ In two separate experiments the pH was measured during the titration of \(5.00\) mmol of each acid with \(0.200 M \mathrm{NaOH}\). Each experiment showed only one stoichiometric point when the data were plotted. In one experiment the stoichiometric point was at \(25.00 \mathrm{~mL}\) added \(\mathrm{NaOH}\), and in the other experiment the stoichiometric point was at \(50.00 \mathrm{~mL} \mathrm{NaOH}\). Explain these results. (See Exercise \(113 .\) )

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

What are the major species in solution after \(\mathrm{NaHSO}_{4}\) is dissolved in water? What happens to the \(\mathrm{pH}\) of the solution as more \(\mathrm{NaHSO}_{4}\) is added? Why? Would the results vary if baking soda \(\left(\mathrm{NaHCO}_{3}\right)\) were used instead?

A certain indicator HIn has a \(\mathrm{p} K_{\mathrm{a}}\) of \(3.00\) and a color change becomes visible when \(7.00 \%\) of the indicator has been converted to \(\mathrm{In}^{-}\). At what \(\mathrm{pH}\) is this color change visible?

When a diprotic acid, \(\mathrm{H}_{2} \mathrm{~A}\), is titrated with \(\mathrm{NaOH}\), the protons on the diprotic acid are generally removed one at a time, resulting in a pH curve that has the following generic shape: a. Notice that the plot has essentially two titration curves. If the first equivalence point occurs at \(100.0 \mathrm{~mL} \mathrm{NaOH}\) added, what volume of \(\mathrm{NaOH}\) added corresponds to the second equivalence point? b. For the following volumes of \(\mathrm{NaOH}\) added, list the major species present after the \(\mathrm{OH}^{-}\) reacts completely. i. \(0 \mathrm{~mL}\) NaOH added ii. between 0 and \(100.0 \mathrm{~mL}\) NaOH added iii. \(100.0 \mathrm{~mL} \mathrm{NaOH}\) added iv. between \(100.0\) and \(200.0 \mathrm{~mL} \mathrm{NaOH}\) added v. \(200.0 \mathrm{~mL} \mathrm{NaOH}\) added vi. after \(200.0 \mathrm{~mL} \mathrm{NaOH}\) added c. If the \(\mathrm{pH}\) at \(50.0 \mathrm{~mL} \mathrm{NaOH}\) added is \(4.0\) and the \(\mathrm{pH}\) at \(150.0 \mathrm{~mL} \mathrm{NaOH}\) added is \(8.0\), determine the values \(K_{\mathrm{a}_{1}}\) and \(K_{\mathrm{a}_{2}}\) for the diprotic acid.
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