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Chapter 14: Problem 104

Consider the titration of \(150.0 \mathrm{mL}\) of \(0.100 \mathrm{M}\) HI by \(0.250 \mathrm{M}\) NaOH. a. Calculate the \(\mathrm{pH}\) after \(20.0 \mathrm{mL}\) of \(\mathrm{NaOH}\) has been added. b. What volume of NaOH must be added so that the \(\mathrm{pH}=\) \(7.00 ?\)

Short Answer

Expert verified
a. The pH after 20.0 mL of NaOH has been added is approximately 1.24. b. 60.0 mL of NaOH must be added to reach a pH of 7.00.

Step by step solution

01

a. Calculate the pH after 20.0 mL of NaOH has been added.

First, we need to calculate the moles of HI and NaOH in the titration. Moles of HI: \(C_{HI} \times V_{HI} = 0.100M \times 150.0mL = 0.0150 \,mol\) Moles of NaOH: \(C_{NaOH} \times V_{NaOH} = 0.250M \times 20.0mL = 0.0050 \,mol\) Now let's calculate the remaining moles of HI after the reaction with 20 mL of NaOH: Moles of HI remaining = 0.0150 mol - 0.0050 mol = 0.0100 mol Now we find the new concentration of HI in the solution: \[C_{HI_{remaining}} = \frac{0.0100 \,mol}{(150.0 + 20.0) mL} = 0.0571 \,M\] Next, we calculate the pH of the solution using the hydrogen ion concentration: \[\mathrm{pH} = -\log_{10}(C_{HI_{remaining}}) = -\log_{10}(0.0571) \approx 1.24\]
02

b. What volume of NaOH must be added so that the pH= 7.00?

At pH = 7.00, the solution is neutral, and moles of H+ and OH- will be equal. We can set up a stoichiometric equation to find the volume of NaOH required. Moles of HI = Moles of NaOH \(C_{HI} \times V_{HI} = C_{NaOH} \times V_{NaOH_{required}}\) Solve for the required volume of NaOH: \(V_{NaOH_{required}} = \frac{C_{HI} \times V_{HI}}{C_{NaOH}} = \frac{0.100 \,M \times 150.0 \,mL}{0.250 \,M} = 60.0\, mL\) So, 60.0 mL of NaOH must be added to reach a pH of 7.00.

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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 \(K_{\mathrm{b}}=1 \times 10^{-10}\).

When a person exercises, muscle contractions produce lactic acid. Moderate increases in lactic acid can be handled by the blood buffers without decreasing the pH of blood. However, excessive amounts of lactic acid can overload the blood buffer system, resulting in a lowering of the blood pH. A condition called acidosis is diagnosed if the blood pH falls to 7.35 or lower. Assume the primary blood buffer system is the carbonate buffer system described in Exercise \(45 .\) Calculate what happens to the \(\left[\mathrm{H}_{2} \mathrm{CO}_{3}\right] /\left[\mathrm{HCO}_{3}^{-}\right]\) ratio in blood when the \(\mathrm{pH}\) decreases from 7.40 to 7.35.

A buffered solution is made by adding \(50.0 \mathrm{g} \mathrm{NH}_{4} \mathrm{Cl}\) to 1.00 \(\mathrm{L}\) of a 0.75-M solution of \(\mathrm{NH}_{3}\). Calculate the pH of the final solution. (Assume no volume change.)

Lactic acid is a common by-product of cellular respiration and is often said to cause the "burn" associated with strenuous activity. A 25.0 -mL sample of 0.100 \(M\) lactic acid (HC \(_{3} \mathrm{H}_{5} \mathrm{O}_{3}\), \(\mathrm{p} K_{\mathrm{a}}=3.86\) is titrated with \(0.100 \mathrm{M}\) NaOH solution. Calculate the \(\mathrm{pH}\) after the addition of \(0.0 \mathrm{mL}, 4.0 \mathrm{mL}, 8.0 \mathrm{mL}, 12.5 \mathrm{mL}\) \(20.0 \mathrm{mL}, 24.0 \mathrm{mL}, 24.5 \mathrm{mL}, 24.9 \mathrm{mL}, 25.0 \mathrm{mL}, 25.1 \mathrm{mL}\) \(26.0 \mathrm{mL}, 28.0 \mathrm{mL},\) and \(30.0 \mathrm{mL}\) of the NaOH. Plot the results of your calculations as pH versus milliliters of NaOH added.

Malonic acid \(\left(\mathrm{HO}_{2} \mathrm{CCH}_{2} \mathrm{CO}_{2} \mathrm{H}\right)\) is a diprotic acid. In the titration of malonic acid with NaOH, stoichiometric points occur at \(\mathrm{pH}=3.9\) and \(8.8 .\) A 25.00 -mL sample of malonic acid of unknown concentration is titrated with 0.0984 \(M \mathrm{NaOH},\) requiring \(31.50 \mathrm{mL}\) of the NaOH solution to reach the phenolphthalein end point. Calculate the concentration of the initial malonic acid solution. (See Exercise \(113 .\) )
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