Question

If 196.0 mL of 0.197 M HCl and 70.0 mL of 0.197 M NaOH are mixed, what is the concentration, in molarity, of the ions in the resulting solution?

Answer 1

First, find the moles each solution contains using this relation:

\[Molarity(M)=\frac{moles _{solute} }{ L _{solution} }\]
L(solution) Molarity(M)
196.0 mL of 0.197 M HCl

70.0 mL of 0.197 M NaOH

One of them is a strong acid and the other a strong base, which will dissociate in water, like this:
\[HCl + H _{2}O <-> H _{3}O ^{+} + Cl ^{-}\]\[NaOH <-> OH^{-} + Na^{+}\]The resulting hydronium (H3O+) and hydroxide ions (OH-) will react with one another, their stoichiometric ratio is 1:1, so 1 mole of HCl will react with 1 mole of NaOH. \[H _{3}O ^{+} + OH^{-} <-> 2H _{2}O\]Find how many moles are left (of either HCl or NaOH) and use it to compute the new molarity using the equation i wrote initially.

\(\color{red}{Take\; into\; account \;the\; volume\; of\; both\; solutions!!!!!}\)

NOTE: i'm assuming they don't want you to include the sodium and chloride ions but if they do, the solubility of of NaCl is very high (35.7 g/100 mL water at 20°C), so you can assume that they're all in solution.

Answer 2

i made a mistake, the arrows in the dissociations for HCl and NaOH should only point towards the right.