Question

answer the following questions related to a reaction between hbr and nickel (ii) hydroxide
a. hbr is classified as a:
strong base
weak acid
weak base
strong acid
part 2 of 3
b. write the balanced equation, including phase symbols, for the reaction between hbr and nickel (ii) hydroxide.
+ → +

Explanation:

Step1: Identify Reactants and Reaction Type

The reaction is between hydrobromic acid ($HBr$) and nickel (II) hydroxide ($Ni(OH)_2$). This is an acid - base (neutralization) reaction, where an acid reacts with a base to form a salt and water. $HBr$ is in aqueous solution ($aq$), and $Ni(OH)_2$ is a solid base (s) (most hydroxides of transition metals are insoluble solids).

Step2: Determine Products

The products of a neutralization reaction between an acid ($HBr$) and a base ($Ni(OH)_2$) are a salt and water. The salt formed will be nickel (II) bromide ($NiBr_2$) and water ($H_2O$). Nickel (II) bromide is soluble in water, so it will be in aqueous solution ($aq$), and water is a liquid ($l$).

Step3: Write Unbalanced Equation

The unbalanced equation is: $HBr(aq)+Ni(OH)_2(s)
ightarrow NiBr_2(aq)+H_2O(l)$

Step4: Balance the Equation

• Balance Br atoms: There are 2 Br atoms in $NiBr_2$, so we put a coefficient of 2 in front of $HBr$: $2HBr(aq)+Ni(OH)_2(s)

ightarrow NiBr_2(aq)+H_2O(l)$

• Balance H and O atoms: Now, on the left side, we have 2 $H$ from $HBr$ and 2 $H$ from $Ni(OH)_2$, total of 4 $H$ atoms, and 2 $O$ atoms from $Ni(OH)_2$. On the right side, in $H_2O$, to get 4 $H$ atoms, we need 2 $H_2O$ molecules (since each $H_2O$ has 2 $H$ atoms). So we put a coefficient of 2 in front of $H_2O$: $2HBr(aq)+Ni(OH)_2(s)

ightarrow NiBr_2(aq)+2H_2O(l)$

Answer:

$2HBr(aq) + Ni(OH)_2(s)
ightarrow NiBr_2(aq) + 2H_2O(l)$