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Chapter 15: Problem 91

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

Short Answer

Expert verified
To achieve a pH of 2.15 in the given solution, approximately 17.5 mL of \( 1.50 \times 10^{-2} M \) NaOH must be added to the 500.0 mL of 0.200 M HCl.

Step by step solution

01

Calculate moles of HCl

To calculate the initial moles of HCl in the solution, use the volume of the solution and the concentration of HCl: moles of HCl = volume × concentration moles of HCl = (500.0 mL) × (0.200 mol/L) Remember to convert mL to L: moles of HCl = (0.500 L) × (0.200 mol/L) = 0.100 mol #Step 2: Calculate the concentration of H_3O^+ ions from the desired pH#
02

Calculate [H_3O^+] from pH

pH is defined as the negative logarithm of the concentration of hydronium ions, H_3O^+. So, we can calculate the concentration of hydronium ions in the final solution using the given pH value: \[ [H_3O^+] = 10^{-pH} \] \[ [H_3O^+] = 10^{-2.15} \] \[ [H_3O^+] = 7.079 \times 10^{-3} \, M \] #Step 3: Calculate moles of H_3O^+ and HCl left in the solution#
03

Calculate moles of H_3O^+ and HCl

Since the target [H_3O^+] in the solution is given, we can find the moles of H_3O^+ in the solution: \[ \text{moles of H}_3\text{O}^{+} = (\text{total volume of solution} ) \times (\text{[H}_3\text{O}^{+}]) \] \[ \text{moles of H}_3\text{O}^{+} = (0.500 L + V_{\text{NaOH}}) \times (7.079 \times 10^{-3}) \] We also know that moles of HCl - moles of NaOH = moles of H_3O^+, Hence, moles of NaOH = moles of HCl - moles of H_3O^+. #Step 4: Calculate moles and volume of NaOH needed#
04

Calculate moles and volume of NaOH

Calculate the moles of NaOH required: moles of NaOH = 0.100 - (0.500 + V_{NaOH}) × (7.079 × 10^{-3} mol/L) Now, we can determine the volume of the NaOH solution: moles of NaOH = (volume of NaOH) × (concentration of NaOH) (0.100 - (0.500 + V_{NaOH}) × (7.079 × 10^{-3})) = V_{NaOH} × (1.50 × 10^{-2} mol/L) Solve for V_{NaOH} to obtain the volume of NaOH required: \[ V_{\text{NaOH}} = \frac{0.100 - (0.500 + V_{\text{NaOH}}) \times (7.079 \times 10^{-3})}{1.50 \times 10^{-2}} \] \[ V_{\text{NaOH}} \approx 0.0175 \, \text{L} \] So, the volume of \( 1.50 \times 10^{-2} M \) NaOH needed to achieve a pH of 2.15 is approximately 0.0175 L or 17.5 mL.

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Most popular questions from this chapter

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?

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

Calculate the pH of each of the following solutions. a. \(0.100 \mathrm{M} \mathrm{HONH}_{2}\left(K_{\mathrm{b}}=1.1 \times 10^{-8}\right)\) b. \(0.100 \mathrm{M} \mathrm{HONH}_{3} \mathrm{Cl}\) c. pure \(\mathrm{H}_{2} \mathrm{O}\) d. a mixture containing \(0.100 \mathrm{M} \mathrm{HONH}_{2}\) and \(0.100 \mathrm{M}\) \(\mathrm{HONH}_{3} \mathrm{Cl}\)

Which of the following mixtures would result in a buffered solution when \(1.0 \mathrm{~L}\) of each of the two solutions are mixed? a. \(0.2 \mathrm{M} \mathrm{HNO}_{3}\) and \(0.4 \mathrm{M} \mathrm{NaNO}_{3}\) b. \(0.2 \mathrm{M} \mathrm{HNO}_{3}\) and \(0.4 \mathrm{M} \mathrm{HF}\) c. \(0.2 \mathrm{M} \mathrm{HNO}_{3}\) and \(0.4 M \mathrm{NaF}\) d. \(0.2 \mathrm{M} \mathrm{HNO}_{3}\) and \(0.4 \mathrm{M} \mathrm{NaOH}\)

Calculate the \(\mathrm{pH}\) at the halfway point and at the equivalence point for each of the following titrations. a. \(100.0 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{HC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}\left(K_{\mathrm{a}}=6.4 \times 10^{-5}\right)\) titrated by \(0.10 \mathrm{M} \mathrm{NaOH}\) b. \(100.0 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\left(K_{\mathrm{b}}=5.6 \times 10^{-4}\right)\) titrated by \(0.20 \mathrm{M} \mathrm{HNO}_{3}\) c. \(100.0 \mathrm{~mL}\) of \(0.50 \mathrm{M} \mathrm{HCl}\) titrated by \(0.25 \mathrm{M} \mathrm{NaOH}\)
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