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Chapter 17: Problem 36

A 20.00 mL sample of \(\mathrm{H}_{3} \mathrm{PO}_{4}(\mathrm{aq})\) requires \(18.67 \mathrm{mL}\) of \(0.1885 \mathrm{M} \mathrm{NaOH}\) for titration from the first to the second equivalence point. What is the molarity of the \(\mathrm{H}_{3} \mathrm{PO}_{4}(\mathrm{aq}) ?\)

Short Answer

Expert verified
The molarity of the \(\mathrm{H}_{3} \mathrm{PO}_{4}(\mathrm{aq})\) solution is 0.0585 M.

Step by step solution

01

Interpreting the problem

It's important to understand molarity M is defined as mol/L. Also, phosphoric acid (\(\mathrm{H}_{3} \mathrm{PO}_{4}(\mathrm{aq})\)) has three hydrogen ions to offer for each molecule, while sodium hydroxide (\(\mathrm{NaOH}\)) can accept only one per molecule. Therefore, 1 mol of \(\mathrm{H}_{3} \mathrm{PO}_{4} (\mathrm{aq})\) will react with 3 mol of \(\mathrm{NaOH}\).
02

Calculating the moles of \(\mathrm{NaOH}\)

To start, calculate the number of moles of \(\mathrm{NaOH}\) added to the solution using its formula \(M = \frac{mol}{L}\). Rearranging this gives \(mol = M * L\). The molarity of the \(\mathrm{NaOH}\) solution is given as 0.1885 M and volume is given as 18.67 mL, which should be converted to liters (0.01867 L). Hence, number of moles becomes 0.1885 * 0.01867 = 0.00352 mol.
03

Calculating the molarity of \(\mathrm{H_{3}PO_{4}}\)

As 1 mol of \(\mathrm{H}_{3} \mathrm{PO}_{4}(\mathrm{aq})\) will react with 3 mol of \(\mathrm{NaOH}\), the number of moles of \(\mathrm{H}_{3} \mathrm{PO}_{4}(\mathrm{aq})\) will be one third of the \(\mathrm{NaOH}\) moles. Hence, the number of moles of \(\mathrm{H}_{3} \mathrm{PO}_{4}(\mathrm{aq})\) = 0.00352 mol / 3 = 0.00117 mol. Finally, molarity can be calculated by dividing the number of moles by the volume (in liters). The volume is 20.00 mL or 0.02 L. Substituting the values, the molarity = 0.00117 mol / 0.02 L = 0.0585 M.

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Most popular questions from this chapter

In the titration of \(25.00 \mathrm{mL}\) of \(0.100 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}\) calculate the number of milliliters of \(0.200 \mathrm{M} \mathrm{NaOH}\) that must be added to reach a pH of (a) \(3.85,\) (b) 5.25 (c) 11.10.

Hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2},\) is a somewhat stronger acid than water. Its ionization is represented by the equation \(\mathrm{H}_{2} \mathrm{O}_{2}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{HO}_{2}^{-}\) In \(1912,\) the following experiments were performed to obtain an approximate value of \(\mathrm{p} K_{\mathrm{a}}\) for this ionization at \(0^{\circ} \mathrm{C} .\) A sample of \(\mathrm{H}_{2} \mathrm{O}_{2}\) was shaken together with a mixture of water and 1 -pentanol. The mixture settled into two layers. At equilibrium, the hydrogen peroxide had distributed itself between the two layers such that the water layer contained 6.78 times as much \(\mathrm{H}_{2} \mathrm{O}_{2}\) as the 1 -pentanol layer. In a second experiment, a sample of \(\mathrm{H}_{2} \mathrm{O}_{2}\) was shaken together with 0.250 M NaOH(aq) and 1-pentanol. At equilibrium, the concentration of \(\mathrm{H}_{2} \mathrm{O}_{2}\) was \(0.00357 \mathrm{M}\) in the 1-pentanol layer and 0.259 M in the aqueous layer. In a third experiment, a sample of \(\mathrm{H}_{2} \mathrm{O}_{2}\) was brought to equilibrium with a mixture of 1 -pentanol and \(0.125 \mathrm{M}\) \(\mathrm{NaOH}(\mathrm{aq}) ;\) the concentrations of the hydrogen peroxide were \(0.00198 \mathrm{M}\) in the 1 -pentanol and \(0.123 \mathrm{M}\) in the aqueous layer. For water at \(0^{\circ} \mathrm{C}, \mathrm{p} K_{\mathrm{w}}=14.94\) Find an approximate value of \(\mathrm{pK}_{\mathrm{a}}\) for \(\mathrm{H}_{2} \mathrm{O}_{2}\) at \(0^{\circ} \mathrm{C}\) [Hint: The hydrogen peroxide concentration in the aqueous layers is the total concentration of \(\mathrm{H}_{2} \mathrm{O}_{2}\) and \(\mathrm{HO}_{2}^{-}\). Assume that the 1 -pentanol solutions contain no ionic species.

Carbonic acid is a weak diprotic acid \(\left(\mathrm{H}_{2} \mathrm{CO}_{3}\right)\) with \(K_{a_{1}}=4.43 \times 10^{-7}\) and \(K_{\mathrm{a}_{2}}=4.73 \times 10^{-11} .\) The equiv- alence points for the titration come at approximately pH 4 and 9. Suitable indicators for use in titrating carbonic acid or carbonate solutions are methyl orange and phenolphthalein. (a) Sketch the titration curve that would be obtained in titrating a sample of \(\mathrm{NaHCO}_{3}(\mathrm{aq})\) with \(1.00 \mathrm{M} \mathrm{HCl}\) (b) Sketch the titration curve for \(\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{aq})\) with 1.00 M HCl. (c) What volume of \(0.100 \mathrm{M} \mathrm{HCl}\) is required for the complete neutralization of \(1.00 \mathrm{g} \mathrm{NaHCO}_{3}(\mathrm{s}) ?\) (d) What volume of \(0.100 \mathrm{M} \mathrm{HCl}\) is required for the complete neutralization of \(1.00 \mathrm{g} \mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{s}) ?\) (e) A sample of NaOH contains a small amount of \(\mathrm{Na}_{2} \mathrm{CO}_{3} .\) For titration to the phenolphthalein end point, \(0.1000 \mathrm{g}\) of this sample requires \(23.98 \mathrm{mL}\) of \(0.1000 \mathrm{M} \mathrm{HCl} .\) An additional \(0.78 \mathrm{mL}\) is required to reach the methyl orange end point. What is the percent \(\mathrm{Na}_{2} \mathrm{CO}_{3},\) by mass, in the sample?

\(\begin{array}{lll}\text { Given } & 1.00 & \mathrm{L}\end{array}\) of a solution that is \(0.100 \mathrm{M}\) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\) and \(0.100 \mathrm{M} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}\) (a) Over what pH range will this solution be an effective buffer? (b) What is the buffer capacity of the solution? That is, how many millimoles of strong acid or strong base can be added to the solution before any significant change in pH occurs?

Determine the following characteristics of the titration curve for \(20.0 \mathrm{mL}\) of \(0.275 \mathrm{M} \mathrm{NH}_{3}(\mathrm{aq})\) titrated with \(0.325 \mathrm{M} \mathrm{HI}(\mathrm{aq})\) (a) the initial \(\mathrm{pH}\) (b) the volume of \(0.325 \mathrm{M} \mathrm{HI}(\mathrm{aq})\) at the equivalence point (c) the \(\mathrm{pH}\) at the half-neutralization point (d) the \(\mathrm{pH}\) at the equivalence point
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