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Chapter 13: Problem 178

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 NaOH.

Short Answer

Expert verified
To ensure that the percent ionization of NH3 is no greater than 0.0010% in a 1.0 L solution of 0.050 M NH3, 0.020 g of NaOH must be added.

Step by step solution

01

Find the concentration of NH3 that corresponds to 0.0010% ionization.

To find the concentration of NH3 that corresponds to 0.0010% ionization, we first write the ionization equation: \[NH_{3(aq)} + H_2O_{(l)} \rightleftharpoons NH_{4(aq)}^{+} + OH_{(aq)}^{-}\] Since we know the initial concentration of NH3 (0.050 M), we can use the percent ionization to find the concentration of NH4+ and OH-: Percent ionization = (Concentration of [OH-]) / (Initial concentration of [NH3]) * 100 0.0010% = (Concentration of [OH-]) / 0.050 Now, solve for the concentration of [OH-]: Concentration of [OH-] = 0.050 * 0.0010% = 5.0 x 10^{-4} M
02

Calculate the molar amount of NaOH to add.

When NaOH is added to the solution, it will dissociate completely into Na+ and OH- ions: \[NaOH_{(s)} \rightarrow Na_{(aq)}^{+} + OH_{(aq)}^{-}\] Since the added NaOH will give an equal amount of OH-, we can determine the amount of NaOH in moles that need to be added to the solution: Moles of NaOH = Molarity of OH- ions * Volume of solution Moles of NaOH = (5.0 x 10^{-4} M) * (1.0 L) = 5.0 x 10^{-4} moles
03

Calculate the mass of NaOH to be added.

Now that we have the molar amount of NaOH, we can determine the mass of NaOH that must be added to the solution: Mass of NaOH = Moles of NaOH * Molar mass of NaOH Molar mass of NaOH = 22.99 g/mol (Na) + 15.99 g/mol (O) + 1.01 g/mol (H) = 39.99 g/mol Mass of NaOH = (5.0 x 10^{-4} moles) * (39.99 g/mol) = 0.020 g Therefore, to ensure that the percent ionization of NH3 is no greater than 0.0010%, 0.020 g of NaOH should be added to the 1.0 L of 0.050 M NH3 solution.

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