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8: Solution Chemistry

  • Page ID
    284426
  • Measuring Concentrations of Solutions

    Exercise \(\PageIndex{1a}\)

    What is the molarity of an NaI solution that contains 4.5 g of NaI in 21.0 mL of solution?

    1. 1.4 M      
    2. 0.030 M  
    3. 0.0047 M     
    4. 0.00014 M       
    5. 0.21 M
    Answer

    a. 1.4 M

     

    \(n_{NaI}=4.5 g NaI (\frac{1 mol}{149.89424})=0.03 mol NaI\)

     

    \(M = \frac{mol}{L}=\frac{0.03}{21*10^{-3}L}=1.429 M\)

    Exercise \(\PageIndex{1b}\)

    What volume of 0.742 M Na2CO3 solution contains 44.9 g of Na2CO3?

    1. 0.314 L
    2. 6.41 × 103 L
    3. 0.571 L
    4. 3.53 × 103 L
    5. 1.75 L
    Answer

    c. 0.571 L

    Exercise \(\PageIndex{1c}\)

    How many moles of sulfate ions are there in a 0.650-L solution of 0.312 M Al2(SO4)3?

    1. 0.203 mol
    2. 0.608 mol
    3. 6.25 mol
    4. 0.0676 mol
    5. 1.44 mol
    Answer

    b. 0.608 mol

    Exercise \(\PageIndex{1d}\)

    What mass of Na2CO3 is present in 0.700 L of a 0.396 M Na2CO3 solution?

    1. 29.4 g
    2. 74.2 g
    3. 42.0 g
    4. 187 g
    5. 60.0 g
    Answer

    a. 29.4 g

    Exercise \(\PageIndex{1e}\)

    In order to dilute 77.1 mL of 0.778 M HCl to 0.100 M, the volume of water that must be added is

    1. 67.2 mL
    2. 9.91 mL
    3. 6 * 102 mL
    4. 1.01 * 10-3 mL
    5. 5.23 * 102 mL
    Answer

    e. 5.23 * 102 mL

     

    \(M_{1}*V_{1}=M_{2}*V_{2}\)

     

    \(V_{2}=\frac{\left (M_{1}*V_{1} \right )}{M_{2}}=\frac{\left ( 0.778 M * \left ( 77.1 * 10^{-3}L \right )\right )}{0.100 M}= .5998 L\)

    Exercise \(\PageIndex{1f}\)

    A dilute solution is prepared by transferring 35.00 mL of a 0.6363 M stock solution to a 900.0 mL volumetric flask and diluting to mark. What is the molarity of this dilute solution?

    1. 0.02474 M
    2. 0.04949 M
    3. 0.1636 M
    4. 0.006186 M
    5. 0.3182 M
    Answer

    a. 0.02474 M

    Exercise \(\PageIndex{1g}\)

    What volume of 1.27 M HCl is required to prepare 197.4 mL of 0.456 M HCl?

    1. 5.5 * 102 mL
    2. 3.41 * 102 mL
    3. 70.9 mL
    4. 0.0141 mL
    5. 1.14 * 102 mL
    Answer

    c. 70.9 mL

     

     

    Acid and Base Concentrations - pH Scale

    Exercise \(\PageIndex{2a}\)

    What is the pH of 4.1 × 10−3 M HCl(aq)? Is this solution acidic or basic?

    Answer

    \(pH = -log(H^{+})=-log(4.1*10^{-3})=2.39\)

    Acidic Solution

    Exercise \(\PageIndex{2b}\)

    What is the pH of a solution that has a H+ concentration of 2.4 * 10-9 M. Is this solution acidic or basic?

    Answer

    pH = 8.62 

    Basic Solution

     

    Exercise \(\PageIndex{2c}\)

    The pH of a vinegar solution is 4.15. What is the H3O+ concentration of the solution?

    Answer

    Note Hydronium ion, H30+, and H+ are all the same thing.

    \(H^{+}=10^{-pH}=10^{-4.15}=7.08 * 10^{-5}M\)

    Exercise \(\PageIndex{2d}\)

    What is the H+ concentration of a solution with a pH of 12.05.

    Answer

    8.91 * 10-13 M

     

     

    Stoichiometry of Aqueous Reactions

    Exercise \(\PageIndex{3a}\)

    The concentration of sulfate in a sample of wastewater is to be determined by using gravimetric analysis. To a 100.0-mL sample of the wastewater is added an excess of barium nitrate, forming the insoluble barium sulfate (233.4 g/mol) according to the balanced equation given below. The solid barium sulfate is dried, and its mass is measured to be 0.9951 g. What was the concentration of sulfate in the original wastewater sample?

    SO42–(aq) + Ba(NO3)2(aq)  →  BaSO4(s) + 2NO3(aq)

    Answer

    \(n_{BaSO_{4}}=\frac{0.9951g BaSO_{4}}{233.4 g/mol}=4.265 * 10^{-3} M\)

    Exercise \(\PageIndex{3b}\)

    Zn reacts with hydrochloric acid.
    Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g)
    What volume of 3.05 M HCl(aq) will react with 25.0 g Zn(s)?

    Answer

    0.251 L

    Exercise \(\PageIndex{3c}\)

    What minimum mass of cobalt (II) nitrate must be added to 60.0 mL of a 0.0999 M phosphate solution in order to completely precipitate all of the phosphate as solid cobalt (II) phosphate?
    2PO43–(aq) + 3Co(NO3)2(aq)  →  Co3(PO4)2(s) + 6NO3(aq)

    Answer

    1.64 g

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