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Chapter 4: Problem 113

A \(15.00-\mathrm{mL}\) solution of potassium nitrate \(\left(\mathrm{KNO}_{3}\right)\) was diluted to \(125.0 \mathrm{~mL}\), and \(25.00 \mathrm{~mL}\) of this solution were then diluted to \(1.000 \times 10^{3} \mathrm{~mL}\). The con- centration of the final solution is \(0.00383 M\). Calculate the concentration of the original solution.

Short Answer

Expert verified
The concentration of the original solution is \(1.276 M\).

Step by step solution

01

Identify given variables

Firstly, there are several variables provided in the problem that are to be identified such as the volume of the original solution \(V1 = 15.00 mL\), dilution volume \(V2 = 125.0 mL\) and the final dilution volume \(V3 = 1000 mL\). Also the molarity of the final solution is given as \(M3 = 0.00383 M\). The molarity of the original solution \(M1\) is what is needed to be calculated.
02

Calculate the molarity of the intermediate solution

This would require using the formula for dilution \(M1V1 = M2V2\). Rearranging the formula to solve for \(M2\), we get \(M2 = M1V1/V2\). However, since \(M1\) is not yet known, the formula can not be directly applied. A second dilution is also mentioned in the problem, hence, it is possible to obtain a relationship between \(M1\), \(M2\) and \(M3\) by taking into account both dilutions which could then be solved for \(M1\). For the second dilution, \(M2V2 = M3V3\), therefore \(M2 = M3V3/V2\). Since \(M3\), \(V3\) and \(V2\) are known, \(M2\) can be calculated, gives \(M2 = (0.00383 M * 1000 mL) / 25.00 mL = 0.15324 M\).
03

Calculate the molarity of the original solution

After calculating the molarity of the intermediate solution, \(M1\) can now be obtained by substituting \(M2\) into the equation \(M1 = M2V2/V1\). Substituting the known values into the equation gives \(M1 = 0.15324 M * 125.0 mL / 15.00 mL = 1.276 M\). Therefore, the molarity of the original solution is \(1.276 M\).

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Most popular questions from this chapter

The molecular formula of malonic acid is \(\mathrm{C}_{3} \mathrm{H}_{4} \mathrm{O}_{4}\). If a solution containing \(0.762 \mathrm{~g}\) of the acid requires \(12.44 \mathrm{~mL}\) of \(1.174 \mathrm{M} \mathrm{NaOH}\) for neutralization, how many ionizable \(\mathrm{H}\) atoms are present in the molecule?

Hydrogen halides (HF, HCl, HBr, HI) are highly reactive compounds that have many industrial and laboratory uses. (a) In the laboratory, HF and \(\mathrm{HCl}\) can be generated by reacting \(\mathrm{CaF}_{2}\) and \(\mathrm{NaCl}\) with concentrated sulfuric acid. Write appropriate equations for the reactions. (Hint: These are not redox reactions.) (b) Why is it that HBr and HI cannot be prepared similarly-that is, by reacting NaBr and NaI with concentrated sulfuric acid? (Hint: \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is a stronger oxidizing agent than both \(\mathrm{Br}_{2}\) and \(\mathrm{I}_{2} .\) ) (c) HBr can be prepared by reacting phosphorus tribromide \(\left(\mathrm{PBr}_{3}\right)\) with water. Write an equation for this reaction.

Calculate the mass of \(\mathrm{KI}\) in grams required to prepare \(5.00 \times 10^{2} \mathrm{~mL}\) of a \(2.80 \mathrm{M}\) solution.

Calculate the concentration (in molarity) of a \(\mathrm{NaOH}\) solution if \(25.0 \mathrm{~mL}\) of the solution are needed to neutralize \(17.4 \mathrm{~mL}\) of a \(0.312 \mathrm{M} \mathrm{HCl}\) solution.

Calculate the molarity of each of the following solutions: (a) \(6.57 \mathrm{~g}\) of methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) in \(1.50 \times\) \(10^{2} \mathrm{~mL}\) of solution, (b) \(10.4 \mathrm{~g}\) of calcium chloride \(\left(\mathrm{CaCl}_{2}\right)\) in \(2.20 \times 10^{2} \mathrm{~mL}\) of solution, (c) \(7.82 \mathrm{~g}\) of naphthalene \(\left(\mathrm{C}_{10} \mathrm{H}_{8}\right)\) in \(85.2 \mathrm{~mL}\) of benzene solution.
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