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

Relevance
• 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{-}

• 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

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

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

• Anonymous
2 months ago

do it