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All aqueous ionic solution are electrically neutral: The number of positive elementary charges is equal to the number of negative elementary charges.
In all aqueous solutions: Number of positive elementary charges = number of negative elementary charges
Two people have the same amount of money, one has $n_1$ 1 Euro coins and $n_2$ 2 Euro coins , the other $n_5$ 5 Euro bills.. So we have: $1$$\cdot n_1\;+\;$$2$$\cdot n_2\;=\; $$5$$\cdot n_5$
1) Aqueous solution of sodium chloride This solution contains the ions $Na^+$ and $Cl^-$ from the dissociation of the salt and the ions $H_3O^+$ and $OH^-$ from the autoprotolysis of water: Electroneutrality condition: $ [Na^+]$ $ +$ $ [H_3O^+]$ $ =$ $ [Cl^-]$ $ + $ $[OH^-] $ 2) Aqueous solution of hydrochloric acid This solution contains ions $H_3O^+$ from the ionization of the acid and the autoprotolysis of the water, the ion $Cl^-$ from the ionization of the acid and the ions $OH^-$ from the autoprotolysis of water: Electroneutrality condition: $ [H_3O^+]$ $ = $ $[Cl^-] $ $+$ $ [OH^-] $ 3) Aqueous solution of calcium bromide This solution contains the ions $Ca^{2+}$ and $Br^-$ from the dissociation of the salt and the ions $H_3O^+$ and $OH^-$ from the autoprotolysis of water: Electroneutrality condition: 2 $[Ca^{2+}] $ $+ $ $[H_3O^+]$ $ = $ $[Br^-]$ $ +$ $ [OH^-] $ 4) Magnesium bromide dissolved in a solution of hydrochloric acid Electroneutrality condition: 2 $[Mg^{2+}] $ $+$ $ [H_3O^+]$ $ = $ $[Br^-]$ $ + $ $[Cl^-]$ $ +$ $ [OH^-] $