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Chapter 14: Problem 112

Consider a solution formed by mixing \(50.0 \mathrm{mL}\) of \(0.100 \mathrm{M}\) \(\mathrm{H}_{2} \mathrm{SO}_{4}, 30.0 \mathrm{mL}\) of \(0.100 \mathrm{M} \mathrm{HOCl}, 25.0 \mathrm{mL}\) of \(0.200 \mathrm{M} \mathrm{NaOH}\), \(25.0 \mathrm{mL}\) of \(0.100 M \mathrm{Ba}(\mathrm{OH})_{2},\) and \(10.0 \mathrm{mL}\) of \(0.150 \mathrm{M} \mathrm{KOH}\) Calculate the pH of this solution.

Short Answer

Expert verified
The pH of the solution is calculated by first finding the moles of each individual species in their respective solutions. Then the net moles of \(H^+\) and \(OH^-\) ions are calculated leading to the determination that the solution will be acidic since the moles of \(OH^-\) ions are less than the moles of \(H^+\) ions. After calculating the total volume of the mixture, we find the net concentration of \(H^+\) ions, which turns out to be 0.09286 M. Finally, using the formula pH = -log[H+], the pH of the solution is found to be approximately 1.032.

Step by step solution

01

Calculate moles of each species

First, let's calculate the moles of each species in their individual solutions.formula is Moles = Molarity × Volume (in L): 1. \(H_2SO_4\) moles: \( 0.100\,\text{M} \times 0.050\,\text{L} = 0.005\,\text{moles} \) Since there are 2 moles of H+ ions in 1 mole of \(H_2SO_4\): moles of H+ ions from \(H_2SO_4\): \( 0.005\,\text{moles} \times 2 = 0.010\,\text{moles} \) 2. \(HOCl\) moles: \( 0.100\,\text{M} \times 0.030\,\text{L} = 0.003\,\text{moles} \) 3. \(NaOH\) moles: \( 0.200\,\text{M} \times 0.025\,\text{L} = 0.005\,\text{moles} \) 4. \(Ba(OH)_2\) moles: \( 0.100\,\text{M} \times 0.025\,\text{L} = 0.0025\,\text{moles} \) Since there are 2 moles of OH- ions in 1 mole of \(Ba(OH)_2\): moles of OH- ions from \(Ba(OH)_2\): \( 0.0025\,\text{moles} \times 2 = 0.005\,\text{moles} \) 5. \(KOH\) moles: \( 0.150\,\text{M} \times 0.010\,\text{L} = 0.0015\,\text{moles} \)
02

Calculate net moles of H+ and OH- ions

Now, let's calculate the net moles of H+ and OH- ions in the mixture: Net moles of H+ ions: \(0.010\,\text{moles} + 0.003\,\text{moles} = 0.013\,\text{moles}\) Net moles of OH- ions: \(0.005\,\text{moles} + 0.005\,\text{moles} + 0.0015\,\text{moles} = 0.0115\,\text{moles}\) Since the moles of OH- ions are less than the moles of H+ ions, we will have an acidic solution.
03

Calculate total volume and H+ concentration

Next, let's calculate the total volume of the mixture: \(0.050\,\text{L} + 0.030\,\text{L } + 0.025\,\text{L} + 0.025\,\text{L} + 0.010\,\text{L} = 0.140\,\text{L}\) Now we can calculate the net concentration of H+ ions, which is: Net H+ moles \(÷\) Total Volume = Net H+ concentration \(0.013\,\text{moles} ÷ 0.140\,\text{L} = 0.09286\,\text{M}\) (rounded to 5 decimal places)
04

Calculate the pH of the solution

Finally, to calculate the pH of the solution, use the formula pH = -log[H+]: pH =\(-\log{(0.09286)}\approx 1.032\) (rounded to 3 decimal places) So, the pH of this solution is approximately 1.032.

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Most popular questions from this chapter

Figure \(14-4\) shows the \(\mathrm{pH}\) curves for the titrations of six different acids by NaOH. Make a similar plot for the titration of three different bases by 0.10 \(M\) HCl. Assume \(50.0 \mathrm{mL}\) of \(0.20 M\) of the bases and assume the three bases are a strong base (KOH), a weak base with \(K_{\mathrm{b}}=1 \times 10^{-5},\) and another weak base with \(K_{\mathrm{b}}=1 \times 10^{-10}\).

You make 1.00 L of a buffered solution \((p H=4.00)\) by mixing acetic acid and sodium acetate. You have \(1.00 M\) solutions of each component of the buffered solution. What volume of each solution do you mix to make such a buffered solution?

Could a buffered solution be made by mixing aqueous solutions of HCl and NaOH? Explain. Why isn't a mixture of a strong acid and its conjugate base considered a buffered solution?

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 \(100.0 \mathrm{mL}\) NaOH added, what volume of NaOH added corresponds to the second equivalence point? b. For the following volumes of NaOH added, list the major species present after the OH \(^{-}\) reacts completely. i. \(0 \mathrm{mL} \mathrm{NaOH}\) added ii. between 0 and \(100.0 \mathrm{mL}\) NaOH added iii. \(100.0 \mathrm{mL}\) NaOH added iv. between 100.0 and \(200.0 \mathrm{mL}\) NaOH added v. \(200.0 \mathrm{mL} \mathrm{NaOH}\) added vi. after \(200.0 \mathrm{mL}\) NaOH added c. If the \(\mathrm{pH}\) at \(50.0 \mathrm{mL}\) 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.

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