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Chapter 4: Problem 77

Pure acetic acid, known as glacial acetic acid, is a liquid with a density of \(1.049 \mathrm{~g} / \mathrm{mL}\) at \(25^{\circ} \mathrm{C}\). Calculate the molarity of a solution of acetic acid made by dissolving \(20.00 \mathrm{~mL}\) of glacial acetic acid at \(25^{\circ} \mathrm{C}\) in enough water to make \(250.0 \mathrm{~mL}\) of solution.

Short Answer

Expert verified
The molarity of the acetic acid solution can be calculated using the following steps: 1. Calculate the mass of acetic acid: \(mass = 1.049 \mathrm{~g/mL} \times 20.00 \mathrm{~mL}\) 2. Calculate moles of acetic acid: \(moles = \frac{mass}{60.05 \mathrm{~g/mol}}\) 3. Calculate molarity of acetic acid in the solution: \(molarity = \frac{moles}{0.2500 \mathrm{~L}}\) After performing the calculations, we will obtain the molarity of the acetic acid solution.

Step by step solution

01

Calculate the mass of acetic acid

Using the density formula, we can find the mass of acetic acid in the solution. Density = \( \frac{mass}{volume} \) So, mass = Density 脳 volume Given the density of glacial acetic acid at \(25^{\circ} \mathrm{C}\) is \(1.049 \mathrm{~g/mL}\) and the volume is \(20.00 \mathrm{~mL}\). Mass of acetic acid = \(1.049 \mathrm{~g/mL} \times 20.00 \mathrm{~mL}\)
02

Calculate moles of acetic acid

To find the moles of acetic acid, we'll use the following formula: Moles = \( \frac{mass}{molar~mass} \) The molar mass of acetic acid (CH鈧僀OOH) is \(12.01 \mathrm{~g/mol} + 4(1.01 \mathrm{~g/mol}) + 2(16.00 \mathrm{~g/mol}) + 1.01 \mathrm{~g/mol} \approx 60.05 \mathrm{~g/mol}\). Using the mass calculated in Step 1, we can find the moles of acetic acid. Moles of acetic acid = \( \frac{mass}{60.05 \mathrm{~g/mol}} \)
03

Calculate molarity of acetic acid in the solution

Now we can calculate the molarity of the acetic acid in the solution using the formula: Molarity = \( \frac{moles}{volume~of~solution} \) The total volume of the solution is given as \(250.0 \mathrm{~mL}\). In order to use this in the molarity formula, we need to convert mL to L. \(250.0 \mathrm{~mL} = 0.2500 \mathrm{~L}\) Now, we have all the necessary data to calculate the molarity: Molarity of acetic acid = \( \frac{moles}{0.2500 \mathrm{~L}} \) By calculating all the values and applying them in the steps above, we will get the molarity of the acetic acid solution.

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Most popular questions from this chapter

Separate samples of a solution of an unknown salt are treated with dilute solutions of \(\mathrm{HBr}, \mathrm{H}_{2} \mathrm{SO}_{4}\), and \(\mathrm{NaOH}\). A precipitate forms in all three cases. Which of the following cations could the solution contain: \(\mathrm{K}^{+} ; \mathrm{Pb}^{2+} ; \mathrm{Ba}^{2+}\) ?

You choose to investigate some of the solubility guidelines for two ions not listed in Table 4.1, the chromate ion \(\left(\mathrm{CrO}_{4}{ }^{2-}\right)\) and the oxalate ion \(\left(\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\right)\). You are given \(0.01 \mathrm{M}\) solutions (A, B, C, D) of four water-soluble salts: $$ \begin{array}{lll} \hline \text { Solution } & \text { Solute } & \text { Color of Solution } \\ \hline \mathrm{A} & \mathrm{Na}_{2} \mathrm{CrO}_{4} & \text { Yellow } \\ \mathrm{B} & \left(\mathrm{NH}_{4}\right)_{2} \mathrm{C}_{2} \mathrm{O}_{4} & \text { Colorless } \\ \mathrm{C} & \mathrm{AgNO}_{3} & \text { Colorless } \\ \mathrm{D} & \mathrm{CaCl}_{2} & \text { Colorless } \\ \hline \end{array} $$ When these solutions are mixed, the following observations are made: $$ \begin{array}{lll} \hline \text { Expt } & \text { Solutions } & \\ \text { Number } & \text { Mixed } & \text { Result } \\ \hline 1 & \mathrm{~A}+\mathrm{B} & \text { No precipitate, yellow solution } \\\ 2 & \mathrm{~A}+\mathrm{C} & \text { Red precipitate forms } \\ 3 & \mathrm{~A}+\mathrm{D} & \text { No precipitate, yellow solution } \\ 4 & \mathrm{~B}+\mathrm{C} & \text { White precipitate forms } \\ 5 & \mathrm{~B}+\mathrm{D} & \text { White precipitate forms } \\ 6 & \mathrm{C}+\mathrm{D} & \text { White precipitate forms } \\ \hline \end{array} $$ (a) Write a net ionic equation for the reaction that occurs in each of the experiments. (b) Identify the precipitate formed, if any, in each of the experiments. (c) Based on these limited observations, which ion tends to form the more soluble salts, chromate or oxalate?

(a) How many milliliters of \(0.120 \mathrm{M} \mathrm{HCl}\) are needed to completely neutralize \(50.0 \mathrm{~mL}\) of \(0.101 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) solution? (b) How many milliliters of \(0.125 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) are needed to neutralize \(0.200 \mathrm{~g}\) of \(\mathrm{NaOH}\) ? (c) If \(55.8 \mathrm{~mL}\) of \(\mathrm{BaCl}_{2}\) solution is needed to precipitate all the sulfate ion in a \(752-\mathrm{mg}\) sample of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\), what is the molarity of the solution? (d) If \(42.7 \mathrm{~mL}\) of \(0.208 \mathrm{M} \mathrm{HCl}\) solution is needed to neutralize a solution of \(\mathrm{Ca}(\mathrm{OH})_{2}\), how many grams of \(\mathrm{Ca}(\mathrm{OH})_{2}\) must be in the solution?

The average adult human male has a total blood volume of \(5.0 \mathrm{~L}\). If the concentration of sodium ion in this average individual is \(0.135 \mathrm{M}\), what is the mass of sodium ion circulating in the blood?

A solid sample of \(\mathrm{Zn}(\mathrm{OH})_{2}\) is added to \(0.350 \mathrm{~L}\) of \(0.500\) \(M\) aqueous HBr. The solution that remains is still acidic. It is then titrated with \(0.500 \mathrm{M} \mathrm{NaOH}\) solution, and it takes \(88.5 \mathrm{~mL}\) of the \(\mathrm{NaOH}\) solution to reach the equivalence point. What mass of \(\mathrm{Zn}(\mathrm{OH})_{2}\) was added to the HBr solution?
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