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Chapter 17: Problem 90

A sample of \(0.2140 \mathrm{~g}\) of an unknown monoprotic acid was dissolved in \(25.0 \mathrm{~mL}\) of water and titrated with \(0.0950 \mathrm{M}\) \(\mathrm{NaOH}\). The acid required \(30.0 \mathrm{~mL}\) of base to reach the equivalence point. (a) What is the molar mass of the acid? (b) After $15.0 \mathrm{~mL}\( of base had been added in the titration, the \)\mathrm{pH}$ was found to be \(6.50 .\) What is the \(K_{a}\) for the unknown acid?

Short Answer

Expert verified
(a) The molar mass of the unknown monoprotic acid is 75.09 g/mol. (b) The Ka value of the unknown acid is 3.16 × 10⁻⁷.

Step by step solution

01

(a) Calculate the moles of NaOH used for titration

To find the moles of NaOH, we will use the concentration and volume of NaOH solution: Moles of NaOH = Concentration of NaOH × Volume of NaOH Moles of NaOH = \(0.0950 \mathrm{M} \times 30.0 \mathrm{~mL}\) Since the volume is in mL, we need to convert it to L by dividing by 1000: Moles of NaOH = \(0.0950 \mathrm{M} \times \frac{30.0 \mathrm{~mL}}{1000}\) Moles of NaOH = 0.00285 moles
02

Calculate the moles of the unknown acid

At the equivalence point, the number of moles of the acid (HA) is equal to the number of moles of the base (NaOH) since they react in a 1:1 ratio. Moles of acid = Moles of NaOH = 0.00285 moles
03

Calculate the molar mass of the unknown acid

To find the molar mass of the acid, we can use the given mass of the sample and the number of moles calculated earlier: Molar mass = \(\frac{Mass}{Moles}\) Molar mass = \(\frac{0.2140 \mathrm{~g}}{0.00285 \mathrm{~mol}}\) Molar mass = 75.09 g/mol Thus, the molar mass of the unknown monoprotic acid is 75.09 g/mol.
04

(b) Calculate the concentration of acid and base at the half-equivalence point

At the half-equivalence point, only half of the NaOH has been added. Thus, the volume of NaOH added is 15 mL, and the moles of NaOH added is: Moles of NaOH = Concentration of NaOH × Volume of NaOH Moles of NaOH = \(0.0950 \mathrm{M} \times \frac{15.0 \mathrm{~mL}}{1000}\) Moles of NaOH = 0.001425 moles Since we are given the pH of the solution at this point (pH = 6.50), we can find the concentration of H₃O⁺ ions: [H₃O⁺] = 10^(-pH) = 10^(-6.50) = 3.16 × 10⁻⁷ M Now, we can determine the concentration of the conjugate base, A⁻, which will be equal the moles of NaOH added: [A⁻] = 0.001425 moles
05

Calculate the concentration of the undissociated acid (HA) at the half-equivalence point

At the half-equivalence point, half of the acid has reacted with the base. The remaining concentration of the undissociated acid is: [HA] = moles of acid - moles of base = 0.00285 - 0.001425 = 0.001425 moles Since the total volume of the solution (acid + NaOH) at the half-equivalence point is 25.0 mL + 15.0 mL = 40.0 mL, we can convert the moles of HA to concentration: [HA] = \(\frac{0.001425 \mathrm{~mol}}{0.040 \mathrm{~L}}\) = 0.0356 M
06

Calculate the Ka value for the unknown acid

Using the concentrations of H₃O⁺, A⁻, and HA, we can calculate the Ka value of the acid: Ka = \(\frac{[H_{3}O^{+}][A^{-}]}{[HA]}\) Ka = \(\frac{(3.16 \times 10^{-7} \mathrm{~M})(0.0356 \mathrm{~M})}{0.0356 \mathrm{~M}}\) Ka = 3.16 × 10⁻⁷ The Ka value of the unknown acid is 3.16 × 10⁻⁷.

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Most popular questions from this chapter

(a) True or false: "solubility" and "solubility-product constant" are the same number for a given compound. (b) Write the expression for the solubility- product constant for each of the following ionic compounds: \(\mathrm{MnCO}_{3}, \mathrm{Hg}(\mathrm{OH})_{2},\) and \(\mathrm{Cu}_{3}\left(\mathrm{PO}_{4}\right)_{2}\).

For each statement, indicate whether it is true or false. (a) The solubility of a slightly soluble salt can be expressed in units of moles per liter. (b) The solubility product of a slightly soluble salt is simply the square of the solubility. (c) The solubility of a slightly soluble salt is independent of the presence of a common ion. (d) The solubility product of a slightly soluble salt is independent of the presence of a common ion.

Which of the following salts will be substantially more soluble in an \(\mathrm{HNO}_{3}\) solution than in pure water: (a) \(\mathrm{BaSO}_{4}\), (b) \(\mathrm{CuS},\) (c) \(\mathrm{Cd}(\mathrm{OH})_{2}\) (d) \(\mathrm{PbF}_{2}\), (e) \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2} ?\)

Consider the equilibrium $$ \mathrm{B}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{HB}^{+}(a q)+\mathrm{OH}^{-}(a q) $$ Suppose that a salt of \(\mathrm{HB}^{+}(a q)\) is added to a solution of \(\mathrm{B}(a q)\) at equilibrium. (a) Will the equilibrium constant for the reaction increase, decrease, or stay the same? (b) Will the concentration of \(\mathrm{B}(a q)\) increase, decrease, or stay the same? (c) Will the pH of the solution increase, decrease, or stay the same?

The value of \(K_{s p}\) for \(\mathrm{Cd}(\mathrm{OH})_{2}\) is $2.5 \times 10^{-14} .(\mathbf{a})$ What is the molar solubility of \(\mathrm{Cd}(\mathrm{OH})_{2} ?(\mathbf{b})\) The solubility of \(\mathrm{Cd}(\mathrm{OH})_{2}\) can be increased through formation of the complex ion \(\mathrm{CdBr}_{4}^{2-}\left(K_{f}=5 \times 10^{3}\right) .\) If solid \(\mathrm{Cd}(\mathrm{OH})_{2}\) is added to a NaBr solution, what is the initial concentration of NaBr needed to increase the molar solubility of \(\mathrm{Cd}(\mathrm{OH})_{2}\) to $1.0 \times 10^{-3} \mathrm{~mol} / \mathrm{L} ?$
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