Skip to main content
Chemistry LibreTexts

4.7 Stoichiometry of Reactions in Aqueous Solutions

  1. Last updated
  2. Save as PDF
  • Page ID
    166326

    • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \( \newcommand{\dsum}{\displaystyle\sum\limits} \)

    \( \newcommand{\dint}{\displaystyle\int\limits} \)

    \( \newcommand{\dlim}{\displaystyle\lim\limits} \)

    \( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

    ( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\id}{\mathrm{id}}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\kernel}{\mathrm{null}\,}\)

    \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\)

    \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\)

    \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    \( \newcommand{\vectorA}[1]{\vec{#1}}      % arrow\)

    \( \newcommand{\vectorAt}[1]{\vec{\text{#1}}}      % arrow\)

    \( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vectorC}[1]{\textbf{#1}} \)

    \( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

    \( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

    \( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \(\newcommand{\longvect}{\overrightarrow}\)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \(\newcommand{\ket}[1]{\left| #1 \right>}\)
    \(\newcommand{\bra}[1]{\left< #1 \right|}\)
    \(\newcommand{\braket}[2]{\left< #1 \vphantom{#2} \right| \left. #2 \vphantom{#1} \right>}\)
    \(\newcommand{\braopket}[3]{\left< #1 \vphantom{#2}\vphantom{#3} \right| #2 \vphantom{#1}\vphantom{#3} \left| #3 \vphantom{#1}\vphantom{#2} \right>}\)
    \(\newcommand{\qmvec}[1]{\mathbf{\vec{#1}}}\)
    \(\newcommand{\op}[1]{\hat{\mathbf{#1}}}\)
    \(\newcommand{\expect}[1]{\langle #1 \rangle}\)
    \(\newcommand{\dfn}[1]{\emph{\textbf{#1}}}\)

    \(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)

    Stoichiometry in Solutions

     

    Exercise \(\PageIndex{1}\)

    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 \(\PageIndex{2}\)

    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 \(\PageIndex{3}\)

    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

     

    Titrations

    Exercise \(\PageIndex{4}\)

    The reaction of HCl with NaOH is represented by the equation

    HCl(aq) + NaOH(aq)  →  NaCl(aq) + H2O(l)

    What volume of 0.614 M HCl is required to titrate 18.8 mL of 0.619 M NaOH?

    Answer

    19.0 mL

    \(V_{HCl}=V_{NaOH}\left ( \frac{M_{NaOH}}{M_{HCl}} \right )=18.8*\left ( \frac{0.619 M}{0.614 M} \right )=18.95 mL\)

     

    Exercise \(\PageIndex{5}\)

    What volume of 0.955 M HCl, in milliliters, is required to titrate 2.152 g of Na2CO3 to the equivalence point?
    Na2CO3(aq) + 2 HCl(aq) → H2O(ℓ) + CO2(g) + 2 NaCl(aq)

    Answer

    42.5 mL HCl

    \(moles Na_{2}CO_{3}=\frac{2.152g}{105.98844g/mol}=0.0203 moles Na_{2}CO_{3}\)

    \[moles HCl = 2*0.0203=0.0406 moles HCl\]

    \[V=\frac{0.0406 moles HCl}{0.955 M}=0.0425 L= 42.5 mL\]

     

    4.7 Stoichiometry of Reactions in Aqueous Solutions is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.

    1. Back to top
    • Was this article helpful?

    Recommended articles

    1. Article type
      Section or Page
      License
      CC BY-NC-SA
      License Version
      4.0
    2. Tags
      This page has no tags.