# 2.E: Basic Tools of Analytical Chemistry (Exercises)

- Page ID
- 5776

1. Indicate how many significant figures are in each of the following numbers.

- 903
- 0.903
- 1.0903
- 0.0903
- 0.09030
- 9.03 × 10
^{2}

2. Round each of the following to three significant figures.

- 0.89377
- 0.89328
- 0.89350
- 0.8997
- 0.08907

3. Round each to the stated number of significant figures.

- the atomic weight of carbon to 4 significant figures
- the atomic weight of oxygen to 3 significant figures
- Avogadro’s number to 4 significant figures
- Faraday’s constant to 3 significant figures

4. Report results for the following calculations to the correct number of significant figures.

- 4.591 + 0.2309 + 67.1 =
- 313 – 273.15 =
- 712 × 8.6 =
- 1.43/0.026 =
- (8.314 × 298)/96485 =
- log(6.53 × 10
^{–5}) = - 10
^{–7.14}= - (6.51 × 10
^{–5}) × (8.14 × 10^{–9}) =

5. A 12.1374 g sample of an ore containing Ni and Co was carried through Fresenius’ analytical scheme shown in Figure 1.1. At point A the combined mass of Ni and Co was found to be 0.2306 g, while at point B the mass of Co was found to be 0.0813 g. Report the weight percent Ni in the ore to the correct number of significant figures.

6. Figure 1.2 shows an analytical method for the analysis of Ni in ores based on the precipitation of Ni^{2}^{+} using dimethylglyoxime. The formula for the precipitate is Ni(C_{4}H_{14}N_{4}O_{4})_{2}. Calculate the precipitate’s formula weight to the correct number of significant figures.

7. An analyst wishes to add 256 mg of Cl^{–} to a reaction mixture. How many mL of 0.217 M BaCl_{2} is this?

8. The concentration of lead in an industrial waste stream is 0.28 ppm. What is its molar concentration?

9. Commercially available concentrated hydrochloric acid is 37.0% w/w HCl. Its density is 1.18 g/mL. Using this information calculate (a) the molarity of concentrated HCl, and (b) the mass and volume (in mL) of solution containing 0.315 moles of HCl.

10. The density of concentrated ammonia, which is 28.0% w/w NH_{3}, is 0.899 g/mL. What volume of this reagent should be diluted to 1.0 × 10^{3} mL to make a solution that is 0.036 M in NH_{3}?

11. A 250.0 mL aqueous solution contains 45.1 μg of a pesticide. Express the pesticide’s concentration in weight percent, in parts per million, and in parts per billion.

12. A city’s water supply is fluoridated by adding NaF. The desired concentration of F^{–} is 1.6 ppm. How many mg of NaF should be added per gallon of treated water if the water supply already is 0.2 ppm in F^{–}?

13. What is the pH of a solution for which the concentration of H^{+} is 6.92 × 10^{–6} M? What is the [H^{+}] in a solution whose pH is 8.923?

14. When using a graduate cylinder, the absolute accuracy with which you can deliver a given volume is ±1% of the cylinder’s maximum volume. What are the absolute and relative uncertainties if you deliver 15 mL of a reagent using a 25 mL graduated cylinder? Repeat for a 50 mL graduated cylinder.

15. Calculate the molarity of a potassium dichromate solution prepared by placing 9.67 grams of K_{2}Cr_{2} O_{7} in a 100-mL volumetric flask, dissolving, and diluting to the calibration mark.

16. For each of the following explain how you would prepare 1.0 L of a solution that is 0.10 M in K^{+}. Repeat for concentrations of 1.0 × 10^{2} ppm K^{+} and 1.0% w/v K^{+}.

- KCl
- K
_{2}SO_{4} - K
_{3}Fe(CN)_{6}

17. A series of dilute NaCl solutions are prepared starting with an initial stock solution of 0.100 M NaCl. Solution A is prepared by pipeting 10 mL of the stock solution into a 250-mL volumetric flask and diluting to volume. Solution B is prepared by pipeting 25 mL of solution A into a 100-mL volumetric flask and diluting to volume. Solution C is prepared by pipeting 20 mL of solution B into a 500-mL volumetric flask and diluting to volume. What is the molar concentration of NaCl in solutions A, B and C?

Note

This is an example of a serial dilution, which is a useful method for preparing very dilute solutions of reagents.

18. Calculate the molar concentration of NaCl, to the correct number of significant figures, if 1.917 g of NaCl is placed in a beaker and dissolved in 50 mL of water measured with a graduated cylinder. If this solution is quantitatively transferred to a 250-mL volumetric flask and diluted to volume, what is its concentration to the correct number of significant figures?

19. What is the molar concentration of NO_{3}^{–}in a solution prepared by mixing 50.0 mL of 0.050 M KNO_{3} with 40.0 mL of 0.075 M NaNO_{3}? What is pNO_{3} for the mixture?

20. What is the molar concentration of Cl^{–} in a solution prepared by mixing 25.0 mL of 0.025 M NaCl with 35.0 mL of 0.050 M BaCl_{2}? What is pCl for the mixture?

21. To determine the concentration of ethanol in cognac a 5.00 mL sample of cognac is diluted to 0.500 L. Analysis of the diluted cognac gives an ethanol concentration of 0.0844 M. What is the molar concentration of ethanol in the undiluted cognac?

### 2.8.2 Solutions to Practice Exercises

#### Practice Exercise 2.1

The correct answer to this exercise is 1.9×10^{–2}. To see why this is correct, let’s work through the problem in a series of steps. Here is the original problem

\[\dfrac{0.250 × (9.93 × 10^{-3}) - 0.100 × (1.927 × 10^{-2})}{9.93 × 10^{-3} + 1.927 × 10^{-2}} =\]

Following the correct order of operations we first complete the two multiplications in the numerator. In each case the answer has three significant figures, although we retain an extra digit, highlighted in red, to avoid round-off errors.

\[\dfrac{2.48{\color{Red} 2} × 10^{-3} - 1.92{\color{Red} 7} × 10^{-3}}{9.93 × 10^{-3} + 1.927 × 10^{-2}} =\]

Completing the subtraction in the numerator leaves us with two significant figures since the last significant digit for each value is in the hundredths place.

\[\dfrac{0.55{\color{Red} 5} × 10^{-3}}{9.93 × 10^{-3} + 1.927 × 10^{-2}} =\]

The two values in the denominator have different exponents. Because we are adding together these values, we first rewrite them using a common exponent.

\[\dfrac{0.55{\color{Red} 5} × 10^{-3}}{0.993 × 10^{-2} + 1.927 × 10^{-2}} =\]

The sum in the denominator has four significant figures since each value has three decimal places.

\[\dfrac{0.55{\color{Red} 5} × 10^{-3}}{2.920 × 10^{-2}} =\]

Finally, we complete the division, which leaves us with a result having two significant figures.

\[\dfrac{0.55{\color{Red} 5} × 10^{-3}}{2.920 × 10^{-2}} = 1.9 × 10^{-2}\]

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#### Practice Exercise 2.2

The concentrations of the two solutions are

\[\mathrm{\dfrac{0.50\: mol\: NaCl}{L} × \dfrac{58.43\: g\: NaCl}{mol\: NaCl} × \dfrac{10^6\: g}{g} × \dfrac{1\: L}{1000\: mL} = 2.9 × 10^3\: g/mL}\]

\[\mathrm{\dfrac{0.25\: mol\: SrCl_2}{L} × \dfrac{158.5\: g\: SrCl_2}{mol\: NaCl} × \dfrac{10^6\: g}{g} × \dfrac{1\: L}{1000\: mL} = 4.0 × 10^3\: g/mL}\]

The solution of SrCl_{2} has the larger concentration when expressed in μg/mL instead of in mol/L.

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#### Practice Exercise 2.3

The concentrations of K^{+} and SO_{4}^{2–} are

\[\mathrm{\dfrac{1.5\: g\: K_2SO_4}{0.500\: L} × \dfrac{1\: mol\: K_2SO_4}{174.3\: g\: K_2SO_4} × \dfrac{2\: mol\: K^+}{mol\: K_2SO_4} = 3.4{\color{Red} 4}×10^{-2}\: M\: K^+}\]

\[\mathrm{\dfrac{1.5\: g\: K_2SO_4}{0.500\: L} × \dfrac{1\: mol\: K_2SO_4}{174.3\: g\: K_2SO_4} × \dfrac{1\: mol\: SO_4^{2-}}{mol\: K_2SO_4} = 1.7{\color{Red} 2}×10^{-2}\: M\: SO_4^{2-}}\]

The pK and pSO_{4} values are

\[\mathrm{pK = -\log(3.4{\color{Red} 4}× 10^{-2}) = 1.46}\]

\[\mathrm{pSO_4 = -\log(1.7{\color{Red} 2} × 10^{-2}) = 1.76}\]

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#### Practice Exercise 2.4

First, we find the moles of AgBr

\[\mathrm{0.250\: g\: AgBr × \dfrac{1\: mol\: AgBr}{187.8\: g\: AgBr} = 1.33{\color{Red} 1}×10^{-3}}\]

and then the moles and volume of Na_{2}S_{2}O_{3}

\[\mathrm{1.33{\color{Red} 1}×10^{-3}\: mol\: AgBr × \dfrac{2\: mol\: Na_2S_2O_3}{mol\: AgBr} = 2.66{\color{Red} 2}×10^{-3}}\]

\[\mathrm{2.66{\color{Red} 2}×10^{-3}\: mol\: Na_2S_2O_3 × \dfrac{1\:L}{0.0138\: mol\: Na_2S_2O_3} × \dfrac{1000\: mL}{L} = 193\: mL}\]

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#### Practice Exercise 2.5

Preparing 500 mL of 0.1250 M KBrO_{3} requires

\[\mathrm{0.500\: L × \dfrac{0.1250\: mol\: KBrO_3}{L} × \dfrac{167.00\: g\: KBrO_3}{mol\: KBrO_3} = 10.44\: g\: KBrO_3}\]

Because the concentration has four significant figures, we must prepare the solution using volumetric glassware. Place a 10.44 g sample of KBrO_{3} in a 500-mL volumetric flask and fill part way with water. Swirl to dissolve the KBrO_{3} and then dilute with water to the flask’s calibration mark.

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#### Practice Exercise 2.6

The first solution is a stock solution, which we then dilute to prepare the standard solution. The concentration of Zn^{2}^{+} in the stock solution is

\[\mathrm{\dfrac{1.004\: g\: Zn}{500\: mL} × \dfrac{10^6\: g}{g} = 2008\: g/mL}\]

To find the concentration of the standard solution we use equation 2.3

\[\mathrm{\dfrac{2008\: g\: Zn^{2+}}{mL} × 2.000\: mL = \mathit{C}_{new}×250.0\: mL}\]

where *C*_{new} is the standard solution’s concentration. Solving gives a concentration of 16.06 μg Zn^{2}^{+}/mL.

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