How do you predict the products in acid-base reactions?

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Answer 1 · 2014-02-19T15:34:11

You predict the products of acid-base reactions by pairing each cation with the anion of the other compound.

Acid-base neutralization is a special case of double displacement reaction.

We can use the same techniques to predict the products.

In acids, the cation is always $"H"^+$ .

Let's use the neutralization of stomach acid ( $"HCl"$ ) by milk of magnesia [ $"Mg(OH)"_2$ ] as an example.

$"HCl + Mg(OH)"_2 → ?$

(a.) Divide the acid into $"H"^+$ ions and negative ions.

$"HCl" → "H"^+ + "Cl"^"-"$

(b.) Divide the base into positive ions and $"OH"^"-"$ ions.

$"Mg(OH)"_2 → "Mg"^"2" + "OH"^"-"$

(c.) For products, combine the $"H"^+$ and $"OH"^"-"$ ions to make $"H"_2"O"$ .

Then combine the positive and negative ions to make a salt.

Remember to balance the positive and negative charges in the salt.

In this case, $"Mg"^"2+"$ and $"Cl"^"-"$ make $"MgCl"_2$ .

$("H"^+ + "Cl"^"-") + ("Mg"^"2+" + "OH"^"-") → ("Mg"^"2+" + "Cl"^"-") + ("H"^+ + "OH"^"-")$

$"HCl + Mg(OH)"_2 → "MgCl"_2 + "H"_2"O"$

(d.) Balance the equation.

$"2HCl" + "Mg(OH)"_2 → "MgCl"_2 + 2"H"_2"O"$

EXAMPLE

Complete and balance an equation for the reaction between $"HBr"$ and $"Al(OH)"_3$ .

Solution

$"HBr" + "Al(OH)"_3 → "?"$

$("H"^+ + "Br"^"-") + ("Al"^"3+" + "OH"^"-") → ("Al"^"3+" + "Br"^"-") + ("H"^+ + "OH"^"-")$

$"HBr + Al(OH)"_3 → "AlBr"_3 +"H"_2"O"$

$"3HBr + Al(OH)"_3 → "AlBr"_3 + "3H"_2"O"$