# Identify each substance as an acid or a base and

Identify each substance as an acid or a base and write a chemical equation showing how it is an acid or a base in aqueous solution according to the Arrhenius definition.
a. NaOH(aq)
b. ${H}_{2}S{O}_{4}\left(aq\right)$.
c. HBr(aq)
d. $Sr{\left(OH\right)}_{2}\left(aq\right)$
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Mary Herrera
Step 1
Part A: $NaOH\left(aq\right)⇒N{a}^{+}\left(aq\right)+O{H}^{-}\left(aq\right)$
NaOH is an Arrhenius base because it produces $O{H}^{-}$ions in an aqueous solution. Write an equation for this ionization reaction.
Step 2
PART B: ${H}_{2}S{O}_{4}\left(aq\right)⇒2{H}^{+}\left(aq\right)+S{O}_{4}^{2-}\left(aq\right)$
${H}_{2}S{O}_{4}$ is an Arrhenius acid because it produces ${H}^{+}$ions in an aqueous solution. Write an equation for this ionization reaction.
Step 3
PART C: $HBr\left(aq\right)⇒{H}^{+}\left(aq\right)+B{r}^{-}\left(aq\right)$
HBr is an Arrhenius acid because it produces ${H}^{+}$ions in an aquation for this ionization reaction.
Step 4
PART D:
$Sr{\left(OH\right)}_{2}\left(aq\right)⇒S{r}^{2+}\left(aq\right)+2O{H}^{-}\left(aq\right)$
$Sr{\left(OH\right)}_{2}$ is an Arrhenius base because it produces $O{H}^{-}$ ions in an aqueous solution. Write an equation for this ionization reaction.

MoxboasteBots5h

(a) Base
(b) Acid
(c) Acid
(d) Base
Explanation:
According to the Arrhenius acid-base theory:
- An acid is a substance that releases ${H}^{+}$ in aqueous solution.
- A base is a substance that releases $O{H}^{-}$ in aqueous solution.
(a) NaOH is a base according to the following equation:
$NaOH\left(aq\right)⇒N{a}^{+}\left(aq\right)+O{H}^{-}\left(aq\right)$
(b) ${H}_{2}S{O}_{4}$ is an acid according to the following equation:
${H}_{2}S{O}_{4}\left(aq\right)⇒2{H}^{+}\left(aq\right)+S{O}_{4}^{2-}\left(aq\right)$
(c) HBr is an acid according to the following equation:
$HBr\left(aq\right)⇒{H}^{+}\left(aq\right)+B{r}^{-}\left(aq\right)$
(d) $Sr{\left(OH\right)}_{2}$ is a base according to the following equation:
$Sr{\left(OH\right)}_{2}\left(aq\right)⇒sS{r}^{2+}\left(aq\right)+2O{H}^{-}\left(aq\right)$

Step 1
Arrhenius theory of acids and based
The Arrhenius theory of acids and bases tells that if a compound generates ${H}^{+}$ ions in aqueous solution then it is identifies as acid whereas if it generates $O{H}^{-}$ ion in aqueous solution then it is identified as a base.
Step 2
1. In water, $NaOH$ dissociate into ions. Since $NaOH\left(aq\right)$ is identified as Arrhenius base.
$NaOH\left(aq\right)⇒N{a}^{+}\left(aq\right)+O{H}^{-}\left(aq\right)$
Step 3
2. In water, ${H}_{2}S{O}_{4}$ dissociate into ions. Since ${H}_{2}S{O}_{4}$ generates ${H}^{+}$ ion into the aqueous solution therefore, ${H}_{2}S{O}_{4}\left(aq\right)$ is identified as Arrhenius acid.
${H}_{2}S{O}_{4}\left(aq\right)⇒2{H}^{+}\left(aq\right)+S{O}_{4}^{2-}\left(aq\right)$
Step 4
3. In water, HBr dissociate into ${H}^{+}$ and $B{r}^{-}$ ions. Since HBr generates ${H}^{+}$ ion into the aqueous solution therefore, HBr(aq) is identified as Arrhenius acid.
$HBr\left(aq\right)⇒{H}^{+}\left(aq\right)+B{r}^{-}\left(aq\right)$
Step 5
4. In water, $Sr\left(OH{\right)}_{2}$ dissociate into $S{r}^{2+}$ and $O{H}^{-}$ ions. Since $Sr\left(OH{\right)}_{2}$ generates $O{H}^{-}$ ion into the aqueous solution therefore, $Sr\left(OH{\right)}_{2}\left(aq\right)$ is identified as Arrhenius base.
$Sr\left(OH{\right)}_{2}\left(aq\right)⇒S{r}^{2+}\left(aq\right)+2O{H}^{-}\left(aq\right)$