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Chapter 4: Problem 73

(a) You have a stock solution of 14.8 \(\mathrm{M} \mathrm{NH}_{3}\) . How many milliliters of this solution should you dilute to make 1000.0 \(\mathrm{mL}\) of 0.250 \(\mathrm{MNH}_{3} ?(\mathbf{b})\) you take a 10.0 -mL portion of the stock solution and dilute it to a total volume of \(0.500 \mathrm{L},\) what will be the concentration of the final solution?

Short Answer

Expert verified
To prepare 1000.0 mL of 0.250 M NH3 solution, you should dilute 16.9 mL of the 14.8 M NH3 stock solution. After diluting a 10.0 mL portion of the stock solution to a total volume of 0.500 L, the final concentration will be 0.296 M NH3.

Step by step solution

01

Identify the given information

Here is the given information for part (a): - Stock solution concentration (M1): 14.8 M NH3 - Desired diluted solution concentration (M2): 0.250 M NH3 - Desired diluted solution volume (V2): 1000.0 mL Our goal is to find the volume of the stock solution needed to make the diluted solution (V1).
02

Use the dilution formula to solve for V1

We will use the dilution formula M1V1 = M2V2 to solve for V1: \(14.8 \times V1 = 0.250 \times 1000.0\) Now we can solve for V1: \(V1 = \frac{0.250 \times 1000.0}{14.8}\)
03

Calculate V1

Plug in the values and calculate V1: \(V1 = \frac{250}{14.8} \approx 16.9 \mathrm{mL}\) So, 16.9 mL of the stock solution should be diluted to make 1000.0 mL of 0.250 M NH3 solution. Part (b):
04

Identify the given information for part (b)

The given information for part (b) is: - Stock solution volume (V1): 10.0 mL - Final diluted solution volume (V2): 0.500 L (we need to convert this to mL) - Stock solution concentration (M1): 14.8 M NH3 Our goal is to find the concentration of the final solution (M2).
05

Convert final volume to mL

We'll convert the final solution volume (V2) from L to mL: V2 = 0.500 L x 1000 = 500 mL
06

Rearrange the dilution formula to solve for M2

Use the dilution formula M1V1 = M2V2 and rearrange it to solve for M2: \(M2 = \frac{M1 \times V1}{V2}\)
07

Calculate M2

Plug in the values and calculate M2: \(M2 = \frac{14.8 \times 10.0}{500} \) M2 = 0.296 M NH3 Hence, the concentration of the final solution after diluting the 10.0 mL stock solution to a total volume of 0.500 L will be 0.296 M NH3.

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Most popular questions from this chapter

The arsenic in a \(1.22-\) g sample of a pesticide was converted to \(\mathrm{AsO}_{4}^{3-}\) by suitable chemical treatment. It was then titrated using \(\mathrm{Ag}^{+}\) to form \(\mathrm{Ag}_{3} \mathrm{AsO}_{4}\) as a precipitate. (a) What is the oxidation state of As in AsO \(_{4}^{3-2}(\mathbf{b})\) Name \(\mathrm{Ag}_{3} \mathrm{AsO}_{4}\) by analogy to the corresponding compound containing phosphorus in place of arsenic. (c) If it took 25.0 \(\mathrm{mL}\) of 0.102 \(\mathrm{MAg}^{+}\) to reach the equivalence point in this titration, what is the mass percentage of arsenic in the pesticide?

Complete and balance the following molecular equations, and then write the net ionic equation for each: \begin{equation} \begin{array}{l}{\text { (a) } \operatorname{HBr}(a q)+\mathrm{Ca}(\mathrm{OH})_{2}(a q) \longrightarrow} \\ {\text { (b) } \mathrm{Cu}(\mathrm{OH})_{2}(s)+\mathrm{HClO}_{4}(a q) \longrightarrow} \\\ {\text { (c) } \mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{HNO}_{3}(a q) \longrightarrow}\end{array} \end{equation}

Which of the following ions will always be a spectator ion in a precipitation reaction? (a) \(\mathrm{Cl}^{-},(\mathbf{b}) \mathrm{NO}_{3}^{-},(\mathbf{c}) \mathrm{NH}_{4}^{+},(\mathbf{d}) \mathrm{S}^{2}\) , (e) \(\mathrm{SO}_{4}^{2-} .\) [Section 4.2\(]\)

The commercial production of nitric acid involves the following chemical reactions: \begin{equation} \begin{array}{c}{4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)} \\ {2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)} \\ {3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(I) \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g)}\end{array} \end{equation} (a) Which of these reactions are redox reactions? (b) In each redox reaction identify the element undergoing oxidation and the element undergoing reduction. (c) How many grams of ammonia must you start with to make 1000.0 L of a 0.150 \(\mathrm{M}\) aqueous solution of nitric acid? Assume all the reactions give 100\(\%\) yield.

In each of the following pairs, indicate which has the higher concentration of \(\mathrm{I}^{-}\) ion: (a) 0.10 \(\mathrm{M}\) BaI \(_{2}\) or 0.25 \(\mathrm{M}\) KI solution, (b) 100 \(\mathrm{mL}\) of 0.10 \(\mathrm{M}\) KI solution or 200 \(\mathrm{mL}\) of 0.040 \(\mathrm{MZnI}_{2}\) solution, \((\mathbf{c}) 3.2 \mathrm{M}\) HI solution or a solution made by dissolving 145 g of Nal in water to make 150 \(\mathrm{mL}\) of solution.
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