$\pu{1 g}$ of $\ce{NaHCO3}$ and $\pu{2 g}$ of $\ce{Na2CO3}$ were dissolved in $\pu{50 mL}$ of water known as solution A. Solution A was mixed with $\pu{450 mL}$ of $\pu{0.154 M}$ of $\ce{NaCl (aq)}$ known as solution C. What is the resulting concentration of $\ce{Cl-}$ ions and $\ce{Na+}$ ions in solution C in $\pu{mM}?$

I started of by finding the amounts of $\ce{NaHCO3}$ and $\ce{Na2CO3}$ which were $1/84$ and $1/53,$ respectively, but I do not know what to do with these numbers.

  • $\begingroup$ General advice: never omit units and solve the problem algebraically first, plug in the physical quantities second. $\endgroup$ – andselisk Oct 16 '20 at 8:22
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    $\begingroup$ @Sarahbaker Using algebraic approach does not have only the aspect of advantage of keeping track of proper approach. The another aspect is the community one. The Questions/Answers are here not just for the OP(i.e. you). Many people search solutions for similar problems all the time. It is of much greater value, if the problem is evaluated generally first in algebraic form. If they have to work just with literal numbers of a particular task, it is much less useful, as it is more difficult to orient in the task and generalize to apply elsewhere $\endgroup$ – Poutnik Oct 16 '20 at 8:43
  • $\begingroup$ @Poutnik It's neither. From Wikipedia: "A physical quantity can be expressed as the combination of a numerical value and a unit. For example, the physical quantity mass can be quantified as n kg, where n is the numerical value and kg is the unit. A physical quantity possesses at least two characteristics in common, one is numerical magnitude and other is the unit in which it is measured." $\endgroup$ – andselisk Oct 16 '20 at 11:09
  • $\begingroup$ @andselisk Sure . But I had in mind rather symbols versus literal instead of [symbols/values] with units or without them. In case of algebraic expresions units are often implicit, as they can vary and can be implied from symbols. Literals require units explicitly as a unit cannot be implied from a number. $\endgroup$ – Poutnik Oct 16 '20 at 11:21

You have 1/84 + 2/106 moles (twice as many sodium ions in a mole of carbonate as in bicarbonate) of sodium ions from the carbonates. Add that to the (0.45 x 0.154) moles of sodium ions from the NaCl. This is your total moles of sodium ions. The total volume of solution is (0.05 + 0.45) L so divide your total moles of sodium ions by 0.5 to get the concentration of sodium ions.

The concentration of Cl- ions is the moles of chloride ions in solution C (0.45 x 0.154) divided by the total volume of solution (0.5L)

  • $\begingroup$ Hey there thanks for replying but did you mean 1/53 rather than 2/53. And for calculating the total moles of sodium, did you mean adding 1/84 + 1/53 + (0.0693). I did not realise you could add moles together $\endgroup$ – Sarah Baker Oct 11 '20 at 13:12
  • $\begingroup$ Edited to correct $\endgroup$ – Waylander Oct 11 '20 at 13:48
  • $\begingroup$ Calling combined amounts of substance "total moles" is as illiterate as calling the combined masses "total grams". Also, current policy suggests the answers should be self-contained and provide straightforward solution. Hints, guidelines and suggestions are better suited as comments. $\endgroup$ – andselisk Oct 16 '20 at 8:29

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