For use in titrations, we want to prepare \(20 \mathrm{L}\) of \(\mathrm{HCl}(\mathrm{aq})\) with a concentration known to four significant figures. This is a two-step procedure beginning with the preparation of a solution of about \(0.10 \mathrm{M}\) HCl. A sample of this dilute HCl(aq) is titrated with a NaOH(aq) solution of known concentration. (a) How many milliliters of concentrated \(\mathrm{HCl}(\mathrm{aq})\) \((d=1.19 \mathrm{g} / \mathrm{mL} ; 38 \% \mathrm{HCl}, \text { by mass })\) must be diluted with water to 20.0 L to prepare \(0.10 \mathrm{M} \mathrm{HCl}\) ? (b) \(\mathrm{A} 25.00\) \(\mathrm{mL}\) sample of the approximately \(0.10\) \(\mathrm{M}\) HCl prepared in part (a) requires \(20.93\) \(\mathrm{mL}\) of \(0.1186\) \(\mathrm{M}\) NaOH for its titration. What is the molarity of the \(\mathrm{HCl}(\mathrm{aq}) ?\) (c) Why is a titration necessary? That is, why not prepare a standard solution of \(0.1000\) \(\mathrm{M} \mathrm{HCl}\) simply by an appropriate dilution of the concentrated HCl(aq)?

Step by step solution

01

Calculate the Volume of Concentrated HCl for Preparation

To prepare a 20.0 L of 0.10 M HCl solution, we will use the formula \( M_{1}V_{1}=M_{2}V_{2} \) for dilution of solutions where\n\( M_{1} \) = Molarity of concentrated HCl (to be found),\n\( V_{1} \) = Volume of concentrated HCl to be added,\n\( M_{2} \) = Molarity of the diluted HCl (0.10 M),\n\( V_{2} \) = Volume of the diluted HCl (20.0 L).\nFrom the given information that the 38% by mass of concentrated HCl solution and \( \text{density (d) is } 1.19 \text{g/mL} \), we calculate \( M_{1} \) as follows:\nThe mass of 1 mL of solution is 1.19 g (from the density). Thus, the mass of HCl in 1 mL of solution will be \( 38\% \) of 1.19 g = 0.4522 g. Now, calculating the moles of HCl in 1 mL of solution using the molar mass of HCl (36.5 g/mol), we find \( M_{1} \) = moles/volume = \( \frac{0.4522 \text{ g}}{36.5 \text{ g/mol}} \) / (1 mL / 1000) = 12.40 mol/L.

02

Substitute and Solve for Unknown

The only unknown value in the dilution equation is \( V_{1} \), the volume of concentrated HCl. Substituting \( M_{1} = 12.4 \text{ M}, M_{2} = 0.1 \text{ M}, V_{2} = 20.0 \text{ L} \) into the dilution formula, we solve for \( V_{1} \): \( V_{1} = \frac{M_{2}V_{2}}{M_{1}} = \frac{(0.1)(20.0)}{12.4} = 0.161 \text{ L} \) which is 161 mL.

03

Determining the Molarity of Dilute HCl after Titration

Following a titration process, 25.00 mL sample of the 0.10 M HCl solution was neutralized by 20.93 mL of 0.1186 M NaOH. Given that HCl and NaOH react in a 1:1 ratio, we can determine the actual molarity of the HCl solution by setting up a mole ratio and solving for unknown molarity. Because the reaction is 1:1, the molarity of HCl will be \( \frac{0.1186 \text{ M NaOH} \times 20.93 \text{ mL NaOH}}{25.00 \text{ mL HCl}} = 0.0991 \text{ M HCl} \)

04

Explanation for the Necessity of Titration

The titration is necessary because the molarity of a solution prepared by dilution isn't always exactly known due to possible experimental errors. Performing titration allows for accurate determination of the solution’s actual concentration, especially when it needs to be known to several significant figures, as in this case. It ensures the precision and reliability of the results.

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