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Chapter 15: Problem 91

Calculate the volume of \(1.50 \times 10^{-2} \mathrm{M} \mathrm{NaOH}\) that must be added to \(500.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{HCl}\) to give a solution that has \(\mathrm{pH}=2.15\)

Short Answer

Expert verified
To find the volume of \(1.50 \times 10^{-2} M\) NaOH solution that must be added to \(500.0 mL\) of \(0.200 M\) HCl to give a solution with a pH of 2.15, we carry out these main steps: (1) Calculate the concentration of H+ ions after addition of NaOH, (2) Calculate the moles of H+ ions, (3) Determine the moles of OH- ions needed to neutralize H+ ions, and (4) Calculate the volume of NaOH solution needed. After completing all steps, we find the volume of NaOH solution required to achieve the desired pH.

Step by step solution

01

Determine the concentration of H+ after adding NaOH

Since the pH is given as 2.15, we can calculate the concentration of H+ ions ([H+]) remaining in the solution after adding NaOH using the formula: \[pH = -\log\text{[H+]}\] Rearrange the formula to find [H+]: \[[H+] = 10^{-pH}\] Substitute the given pH value, and calculate the remaining concentration of H+ ions.
02

Calculate the moles of H+ ions

To calculate the volume of NaOH required to reach the desired pH, we first need to find the moles of H+ ions in the final solution. We can use the formula: \[moles\;of\;H+ = [H+]\times V_{final}\] Where the final volume of the solution, \(V_{final}\), is the sum of the initial volume of HCl solution and the volume of NaOH added. Note that since we don't know the volume of NaOH added yet, let's denote it as V_NaOH. Now we can write the equation: \[moles\;of\;H+ = [H+]\times(V_{initial}+V_{NaOH})\] Use the values of [H+] calculated in Step 1, and given initial volume of the HCl solution, 500.0 mL, to calculate moles of H+.
03

Calculate moles of OH- ions needed to neutralize H+ ions

To increase pH, we need to neutralize some of the H+ ions present, by adding the required amount of NaOH. The balanced chemical equation for the neutralization is: \[\mathrm{HCl + NaOH \rightarrow NaCl + H_2O}\] This shows that one mole of HCl reacts with one mole of NaOH to produce one mole of water and one mole of NaCl. Knowing the moles of H+ ions from Step 2, we can now calculate the moles of OH- ions needed. As the molar ratio between HCl and NaOH in the reaction is 1:1, the moles of OH- required to react with the moles of H+ ions is equal to the calculated moles of H+.
04

Calculate the volume of NaOH needed

Now, we can calculate the volume of NaOH needed to provide the required moles of OH- ions. We can use the formula: \[V_{NaOH} = \frac{moles\;of\;OH-}{[\mathrm{OH^-}]}\] where [\(\mathrm{OH^-}\)] is the concentration of the NaOH solution, and moles of OH- is the amount we calculated in Step 3. Use the given value of NaOH concentration, \(1.50 \times 10^{-2} M\), to calculate the volume of the NaOH solution. In conclusion, the calculated volume of NaOH solution in step 4 needs to be added to the 500.0 mL of 0.200 M HCl solution to achieve a pH of 2.15.

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Most popular questions from this chapter

Repeat the procedure in Exercise 55 , 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 \(0.100 \mathrm{M} \mathrm{HCl}\).

Lactic acid is a common by-product of cellular respiration and is often said to cause the "burn" associated with strenuous activity. A \(25.0-\mathrm{mL}\) sample of \(0.100 M\) lactic acid \(\left(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{3}, \mathrm{p} K_{\mathrm{a}}=\right.\) \(3.86\) ) is titrated with \(0.100 M\) 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 \(\mathrm{NaOH}\). Plot the results of your calculations as pH versus milliliters of \(\mathrm{NaOH}\) added.

A \(0.400 M\) solution of ammonia was titrated with hydrochloric acid to the equivalence point, where the total volume was \(1.50\) times the original volume. At what \(\mathrm{pH}\) does the equivalence point occur?

A student titrates an unknown weak acid, HA, to a pale pink phenolphthalein end point with \(25.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{NaOH}\). The student then adds \(13.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{HCl}\). The \(\mathrm{pH}\) of the resulting solution is \(4.7\). How is the value of \(\mathrm{p} K_{\mathrm{a}}\) for the unknown acid related to \(4.7 ?\)

A buffer solution is prepared by mixing \(75.0 \mathrm{~mL}\) of \(0.275 \mathrm{M}\) fluorobenzoic acid \(\left(\mathrm{C}_{7} \mathrm{H}_{3} \mathrm{O}_{2} \mathrm{~F}\right)\) with \(55.0 \mathrm{~mL}\) of \(0.472 \mathrm{M}\) sodium fluorobenzoate. The \(\mathrm{p} K_{\mathrm{a}}\) of this weak acid is \(2.90\). What is the \(\mathrm{pH}\) of the buffer solution?
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