Solution Stoichiometry溶液计量

Big Idea · Scale, Proportion, and Quantity (SPQ)AP Science Practice 5 · Mathematical RoutinesAP link · Unit 4~45 min

Learning goals

  • Use moles = M × V to convert between solution volume, molarity, and moles in stoichiometry problems
  • Find the mass of precipitate formed when two solutions are mixed, and identify the limiting and excess reactants
  • Read a buret and use titration data to find the concentration of an unknown solution

Before you start

  • Stoichiometry (mole ratios) (2.6.5)
  • Ionic and net ionic equations (2.7.4)
  • Molarity (2.9.2)

Exactly how much base?

In 1.8.5 you used phenolphthalein to tell when NaOH had just neutralized hydrochloric acid. Now the question is how much. If you have hydrochloric acid of unknown concentration, you can neutralize it with NaOH solution of known molarity and record how many milliliters it took. That tells you the acid’s concentration. This method is called titration, and it rests on the solution stoichiometry in this lesson.

The mole bridge for solutions

The “mole bridge” from 2.6.5 runs from the mass of one substance, to its moles, across the mole ratio to moles of another substance, and back to mass. In 2.8.6 you added gas volumes at either end. For solutions you need one more conversion:

mol=M×V\text{mol} = M \times V
  1. Convert what you’re given to moles: M × V for a solution, mass ÷ molar mass for a solid, and V ÷ 22.4 L/mol for a gas at STP.
  2. Use the mole ratio from the balanced equation to find moles of the substance you want.
  3. Convert to what you need: volume of solution = mol ÷ M, molarity = mol ÷ L, or mass = mol × molar mass.

Reactions in solution are really reactions between ions (2.7.4), so you can use either the full equation or the net ionic equation, as long as the mole ratio is right.

Pick a reaction and try finding the mass of precipitate from a solution, or the volume of a second solution needed. Then switch to “Mix two solutions” to see which reactant runs out:

Try it: solution stoichiometry (M × V ⇄ moles ⇄ mass)
Given:
Find:

    Mixing two solutions: which runs out?

    When two solutions are mixed, find the moles of each reactant first, then use the mole ratio to find the limiting reactant限量试剂 (2.6.6). It determines how much product forms, and the excess reactant is left over in solution.

    Acid–base titration

    In an acid–base titration酸碱中和滴定, you add a solution of known concentration from a buret to a measured volume of an unknown acid or base until the reaction is just complete. The solution in the buret is the titrant滴定剂, often a standard solution标准溶液 whose molarity is known precisely. The solution in the flask is the analyte待测液.

    • Reading a buret: 0.00 mL is at the top, and the numbers increase downward. The smallest marks are 0.1 mL, so you estimate one more digit, to 0.01 mL. Volume delivered = final reading − initial reading.
    • Indicator: phenolphthalein is colorless in acid and pink in base. Each drop of NaOH makes a pink swirl that fades as you mix. The end point滴定终点 is reached when the whole solution turns pale pink and stays pink for about 30 seconds.
    • Equivalence point: the point where the moles of base added exactly match the acid by the mole ratio is the equivalence point化学计量点. A good indicator changes color very close to it.
    • Calculation: for HCl and NaOH, which react 1 : 1,
    Macid=MNaOH×VNaOHVacidM_\text{acid} = \frac{M_\text{NaOH} \times V_\text{NaOH}}{V_\text{acid}}

    A buret’s scale runs the opposite way from a graduated cylinder’s, so looking down or up makes the opposite error. Drag the slider to try it:

    Try it: pick an instrument and drag the slider to move your eye
    2122232425↑ 0 markat the top
    Looking upEye levelLooking down

    RuleA buret’s 0 mark is at the top. Looking down gives a reading that’s too low, and looking up gives one that’s too high: the opposite of a graduated cylinder.

    WhyA buret’s scale increases downward. Looking down, your line of sight crosses the wall above the bottom of the meniscus, where the numbers are smaller.

    Run a titration yourself. Add 5 mL or 1 mL at a time at first; when the pink starts fading slowly, switch to one drop at a time. At the end point, read the final volume on the magnified scale and calculate the acid’s molarity:

    Try it: acid–base titration
    Acid (analyte)

    magnified scale
    Add

    Initial reading
    Final reading mL

      This lesson uses the atomic masses on the AP periodic table (H 1.008, O 16.00, Na 22.99, S 32.06, Cl 35.45, Cu 63.55, Ag 107.87, Ba 137.33).

      How Chinese textbooks differ

      • Chinese textbooks write the bridge as n = cV, with c in mol/L, which is the same thing as mol = M × V.
      • At this level, Chinese textbooks speak only of the 滴定终点 (end point). The difference between end point and equivalence point, and choosing an indicator to match, come later in the Chinese course (选择性必修 1, chapter 3), together with titration curves.
      • Chinese problems often ask about reading errors, such as looking down at the buret at the end. Readings increase downward, so a reading taken from above is too small.

      Example 1: how much silver chloride?

      You mix 50.0 mL of 0.100 M AgNO₃ with 20.0 mL of 0.200 M NaCl. What mass of AgCl precipitates? Which reactant is in excess, and how much of it is left?

      AgX+(aq)+ClX−(aq)→AgCl(s)\ce{Ag+(aq) + Cl-(aq) -> AgCl(s)}
      Step-by-step solution

      1. Moles of each. mol = MV, with V in liters:

        n(AgNOX3)=0.100×0.0500=5.00×10−3 moln(NaCl)=0.200×0.0200=4.00×10−3 mol\begin{aligned} n(\ce{AgNO3}) &= 0.100 \times 0.0500 \\ &= 5.00 \times 10^{-3}\ \text{mol} \\ n(\ce{NaCl}) &= 0.200 \times 0.0200 \\ &= 4.00 \times 10^{-3}\ \text{mol} \end{aligned}

      2. Limiting reactant. Ag⁺ and Cl⁻ react 1 : 1. There is less Cl⁻, so NaCl is limiting and AgNO₃ is in excess. The limiting reactant sets the amount of product.

      3. Mass of precipitate. mol AgCl = mol NaCl = 4.00 × 10⁻³ mol, and the molar mass of AgCl is 107.87 + 35.45 = 143.32 g/mol:

        m=4.00×10−3×143.32=0.573 g\begin{aligned} m &= 4.00 \times 10^{-3} \times 143.32 \\ &= \boxed{0.573\ \text{g}} \end{aligned}

      4. Excess AgNO₃ left. The AgNO₃ used equals the NaCl used:

        n=5.00×10−3−4.00×10−3=1.00×10−3 mol\begin{aligned} n &= 5.00 \times 10^{-3} - 4.00 \times 10^{-3} \\ &= \boxed{1.00 \times 10^{-3}\ \text{mol}} \end{aligned}

      5. One step further. The mixture is about 70.0 mL (assuming the volumes add), so the leftover [Ag⁺] is about 1.00 × 10⁻³ mol ÷ 0.0700 L = 0.0143 M.

      6. A common mistake. Comparing molarities: 0.200 M > 0.100 M, so NaCl must be in excess. Compare moles, which include the volumes.

      Example 2: titrating sulfuric acid

      You titrate 25.00 mL of sulfuric acid with 0.1000 M NaOH, using phenolphthalein. The initial buret reading is 0.50 mL and the final reading is 24.30 mL. What is the molarity of the sulfuric acid?

      HX2SOX4(aq)+2 NaOH(aq)→NaX2SOX4(aq)+2 HX2O(l)\ce{H2SO4(aq) + 2NaOH(aq) -> Na2SO4(aq) + 2H2O(l)}
      Step-by-step solution

      1. Volume of NaOH. The volume delivered is final minus initial, read to 0.01 mL:

        V=24.30−0.50=23.80 mLV = 24.30 - 0.50 = 23.80\ \text{mL}

      2. Moles of NaOH.

        n(NaOH)=0.1000×0.02380=2.380×10−3 mol\begin{aligned} n(\ce{NaOH}) &= 0.1000 \times 0.02380 \\ &= 2.380 \times 10^{-3}\ \text{mol} \end{aligned}

      3. Mole ratio. H₂SO₄ and NaOH react 1 : 2:

        n(HX2SOX4)=12×2.380×10−3=1.190×10−3 mol\begin{aligned} n(\ce{H2SO4}) &= \tfrac{1}{2} \times 2.380 \times 10^{-3} \\ &= 1.190 \times 10^{-3}\ \text{mol} \end{aligned}

      4. Molarity.

        [HX2SOX4]=1.190×10−30.02500=0.04760 M\begin{aligned} [\ce{H2SO4}] &= \frac{1.190 \times 10^{-3}}{0.02500} \\ &= \boxed{0.04760\ \text{M}} \end{aligned}

      5. Check. Slightly less NaOH solution was used than acid, and H₂SO₄ is diprotic, so the acid should be a bit less than half as concentrated as the NaOH (0.05 M). 0.04760 M is reasonable.

      6. A common mistake. Using the final reading, 24.30 mL, as the volume delivered, or forgetting the 1 : 2 ratio and getting 0.09520 M.

      Practice: what volume is needed for complete reaction, or what is the molarity?
      1. Volume of 0.100 M NaOH to neutralize 20.0 mL of 0.100 M HCl
      2. Volume of 0.100 M NaOH to neutralize 20.0 mL of 0.100 M H₂SO₄
      3. Volume of 0.400 M NaOH to neutralize 20.0 mL of 0.200 M HCl
      4. 25.0 mL of HCl is neutralized by 25.0 mL of 0.100 M NaOH: [HCl]
      5. 20.0 mL of H₂SO₄ is neutralized by 20.0 mL of 0.100 M NaOH: [H₂SO₄]
      6. Volume of 0.100 M Na₂SO₄ to precipitate all the Ba²⁺ in 10.0 mL of 0.200 M BaCl₂
      7. 10.0 mL of NaOH is neutralized by 20.0 mL of 0.100 M HCl: [NaOH]
      8. Volume of 0.100 M NaOH to precipitate all the Cu²⁺ in 20.0 mL of 0.100 M CuSO₄ as Cu(OH)₂

      Common mistakes

      Mistake 1: “The final buret reading is the volume used.” The initial reading usually isn’t zero, so the volume used is final − initial. Remember too that the scale reads 0 at the top and increases downward.

      Mistake 2: “At neutralization, moles of acid equal moles of base.” Use the mole ratio. H₂SO₄ and NaOH react 1 : 2, so you need twice as many moles of NaOH. What’s equal is the moles of H⁺ and OH⁻.

      Mistake 3: “Stop as soon as you see pink.” A pink swirl that fades means acid is still left; the end point is when the pale pink lasts for 30 seconds. If the solution turns dark pink, you’ve overshot, and the calculated acid concentration will be too high.

      Quick check

      10.0 mL of 0.100 M BaCl₂ reacts with excess Na₂SO₄. What mass of BaSO₄ precipitates?

      A student titrates HCl with standard NaOH but reads the final buret volume with their eye above the meniscus. The calculated [HCl] will be:

      Summary

      • Solution stoichiometry adds mol = M × V to the ends of the mole bridge: given amount → moles → mole ratio → amount wanted.
      • When two solutions are mixed, compare moles and use the limiting reactant to find the product.
      • In a titration, a standard solution reveals an unknown concentration. Read the buret to 0.01 mL, use final − initial, and stop at the first pale pink that lasts 30 seconds.
      On to AP (for later)

      AP topic 4.5 often starts stoichiometry from a solution (M × V), and topic 4.6 introduces titration: finding an unknown concentration from titration data, or finding how much of an ion was in a solution from the mass of a precipitate (gravimetric analysis). Topic 8.5 returns to titration curves, using them to locate the equivalence point and choose an indicator. Lab questions often ask which error would make a calculated concentration too high or too low.

      Practice this chapter: 2.9 · Solutions · 10 questions

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