Anonymous
Anonymous asked in Science & MathematicsChemistry · 2 months ago

Determine the molar concentration of each ion present in the solutions that result from each of the following mixtures:?

(a) 57.8 mL of 0.87 M NaCl and 80.7 mL of 1.74 M NaCl

(b) 111 mL of 0.85 M HCl and 111 mL of 0.10 M HCl(c) 39.1 mL of 0.304 M Ba(NO3)2 and 21.4 mL of 0.261 M AgNO3

(d) 11.6 mL of 0.592 M NaCl and 23.6 mL of 0.176 M Ca(C2H3O2)2

3 Answers

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  • 2 months ago

    Supposing additive volumes and complete ionization in all cases:

    (a)

    ((57.8 mL x 0.87 M) + (80.7 mL x 1.74 M)) / (57.8 mL + 80.7 mL) = 1.38 M NaCl

    So the concentration of ions would be 1.38 M Na+ and 1.38 M Cl-.

    (b)

    ((111 mL x 0.85 M) + (111 mL x 0.10 M)) / (111 mL + 111 mL) = 0.475 M HCl

    Ions: 0.475 M H+ and 0.475 M Cl-

    (c)

    (39.1 mL) x (0.304 M Ba(NO3)2) / (39.1 mL + 21.4 mL) = 0.19646 M Ba(NO3)2

    (21.4 mL) x (0.261 M AgNO3) / (39.1 mL + 21.4 mL) = 0.092321 M AgNO3

    Ba(NO3)2 → Ba{2+} + 2 NO3{-}

    AgNO3 → Ag{+} + NO3{-}

    Ions:

    (0.19646 M Ba(NO3)2) x (1 mol Ba{2+} / 1 mol Ba(NO3)2) = 0.196 M Ba{2+}

    (0.092321 M AgNO3) x (1 mol Ag{+} / 1 mol AgNO3) = 0.0923 M Ag{+}

    ((0.19646 M Ba(NO3)2) x (2 mol NO3- / 1 mol Ba(NO3)2)) +

    ((0.092321 M AgNO3) x (1 mol NO3- / 1 mol AgNO3)) = 0.485 M NO3{-}

    (d)

    (11.6 mL x 0.592 M NaCl) / (11.6 mL + 23.6 mL) = 0.19509 M NaCl

    (23.6 mL x 0.176 M Ca(C2H3O2)2) / (11.6 mL + 23.6 mL) =

    0.11800 M Ca(C2H3O2)2

    NaCl → Na{+} + Cl{-}

    Ca(C2H3O2)2 → Ca{2+} + 2 C2H3O2{-}

    Ions:

    (0.19509 M NaCl) x (1 mol Na{+} / 1 mol NaCl) = 0.195 M Na{+}

    (0.19509 M NaCl) x (1 mol Cl{-} / 1 mol NaCl) = 0.195 M Cl{-}

    (0.11800 M Ca(C2H3O2)2) x (1 mol Ca{+} / 1 mol Ca(C2H3O2)2) = 0.118 M Ca{2+}

    (0.11800 M Ca(C2H3O2)2) x (2 mol C2H3O2{-} / 1 mol Ca(C2H3O2)2) = 

    0.236 M C2H3O2{-}

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  • david
    Lv 7
    2 months ago

    I'll do one ... use it as an example then do the rest.

     ...   (a) 57.8 mL of 0.87 M NaCl and 80.7 mL of 1.74 M NaCl

       ... step one find MOLES of each substance

       moles = molarity X vol. in Liters

       57.8 mL of 0.87 M NaCl 

       moles Na(+) = 0.87 X 57.8 / 1000  =  0.0503 moles

       moles Cl(-) = 0.87 X 57.8 / 1000 = 0.0503 moles

    ========================================

      80.7 mL of 1.74 M NaCl

       moles Na(+) = 1.74 X 80.7 / 1000 = 0.140 moles

       moles Cl(-) = 1.74 X 80.7 / 1000 = 0.140 moles

    ========================================

     ... add moles of same ions

      Na(+) =  0.0503 + 0.140  =  0.1903 moles 

      Cl(-) = 0.0503 + 0.140 = 0.1903 moles 

    ==================================

    add to get total volume

       57.8/1000  +  80.7/1000  =  0.1385 Liters

    ===================================

    last ..  find molarity

      Na(+)  =  0.1903/0.1385  =  1.37 M

      Cl(-)  =  0.1903/0.1385 = 1.37 M

    ==================================

    above are the answers ... Even though these are the same, all will NOT be.. but the same method is used for all ...  Good luck

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  • Anonymous
    2 months ago

    do it          

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