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Chapter 14: Problem 59

How would you prepare \(1600 \mathrm{~mL}\) of a \(\mathrm{pH}=1.50\) solution using concentrated \((12 M) \mathrm{HCl} ?\)

Short Answer

Expert verified
To prepare 1600 mL of a pH 1.50 solution using concentrated 12 M HCl, mix 4.213 mL of concentrated HCl with 1595.787 mL of water.

Step by step solution

01

Calculate the concentration of H+ ions required for the desired pH

: We can use the pH formula to determine the concentration of H+ ions in the solution: \[ pH = -\log[H^+] \] Where \(pH\) is the desired pH (1.50) and \([H^+]\) is the required concentration of H+ ions. Rearrange the formula and solve for \([H^+]\): \[ [H^+] = 10^{-pH} \] Plug in the desired pH (1.50) and calculate the required H+ concentration: \[ [H^+] = 10^{-1.50} \] \[ [H^+] = 0.0316 \ \mathrm{M} \]
02

Calculate the moles of H+ ions required to achieve the desired concentration

: Now that we have the concentration of H+ ions required, we can calculate the moles of H+ ions needed for the entire solution. Use the formula: \[ moles\ of\ H^+ = [H^+] \times \ volume\ of\ solution \] Plug in the H+ concentration (0.0316 M) and the volume of solution (1600 mL, converting to L for the calculation): \[ moles\ of\ H^+ = 0.0316\ \mathrm{M} \times 1.6\ \mathrm{L} \] \[ moles\ of\ H^+ = 0.05056 \]
03

Calculate the volume of 12 M HCl needed to provide the required number of H+ moles

: Now we will determine the volume of 12 M HCl needed to provide the required number of moles of H+ ions. Use the formula: \[ volume\ of\ HCl = \frac{moles\ of\ H^+}{concentration\ of\ HCl} \] Plug in the moles of H+ (0.05056) and the concentration of HCl (12 M): \[ volume\ of\ HCl = \frac{0.05056}{12\ \mathrm{M}} \] \[ volume\ of\ HCl = 0.004213\ \mathrm{L} \] Convert the volume to milliliters for practical use: \[ volume\ of\ HCl = 4.213\ \mathrm{mL} \]
04

Mix the calculated volume of concentrated HCl with the appropriate amount of water

: Finally, mix the calculated volume of concentrated HCl (4.213 mL) with the appropriate amount of water to achieve a total volume of 1600 mL. Subtract the HCl volume from the desired total volume to find the volume of water needed: \[ volume\ of\ water = 1600\ \mathrm{mL} - 4.213\ \mathrm{mL}\] \[ volume\ of\ water = 1595.787\ \mathrm{mL} \] To prepare 1600 mL of a pH 1.50 solution using concentrated 12 M HCl, mix 4.213 mL of concentrated HCl with 1595.787 mL of water.

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Most popular questions from this chapter

Calculate the \(\mathrm{pH}\) of each of the following solutions. a. \(0.12 \mathrm{M} \mathrm{KNO}_{2}\) c. \(0.40 \mathrm{M} \mathrm{NH}_{4} \mathrm{ClO}_{4}\) b. \(0.45 \mathrm{M} \mathrm{NaOCl}\)

What mass of \(\mathrm{NaOH}(s)\) must be added to \(1.0 \mathrm{~L}\) of \(0.050 \mathrm{M} \mathrm{NH}_{3}\) to ensure that the percent ionization of \(\mathrm{NH}_{3}\) is no greater than \(0.0010 \%\) ? Assume no volume change on addition of \(\mathrm{NaOH}\).

Calculate the \(\mathrm{pH}\) of a \(0.200 \mathrm{M}\) solution of \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NHF}\). Hint: \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NHF}\) is a salt composed of \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}\) and \(\mathrm{F}^{-}\) ions. The prin- cipal equilibrium in this solution is the best acid reacting with the best base; the reaction for the principal equilibrium is \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons\) \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}(a q)+\mathrm{HF}(a q) \quad K=8.2 \times 10^{-3}\)

The \(\mathrm{pH}\) of a sample of gastric juice in a person's stomach is \(2.1\). Calculate the \(\mathrm{pOH},\left[\mathrm{H}^{+}\right]\), and \(\left[\mathrm{OH}^{-}\right]\) for this sample. Is gastric juice acidic or basic?

Arrange the following \(0.10 M\) solutions in order from most acidic to most basic. See Appendix 5 for \(K_{\mathrm{a}}\) and \(K_{\mathrm{b}}\) values. \(\mathrm{CaBr}_{2}, \quad \mathrm{KNO}_{2}, \quad \mathrm{HClO}_{4}, \quad \mathrm{HNO}_{2}, \quad \mathrm{HONH}_{3} \mathrm{ClO}_{4}\).
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