Question

The [H3O+] in a 0.050 M solution of Ba(OH)2 is

Answer 1

Answer:

Explanation:

1. 2.0 × 10−5 M

2. 5.0 × 10−10 M

3. 5.0 × 10−2 M

4. 1.0 × 10−13 M

5. 1.0 × 10−5 M

6. 2.0 × 10−13 M

Answer 2

Answer:

The concentration of  $H_{3} O^{+}$ in 0.05M  $Ba(OH)_{2}$ solution is $10^{-13}M$.

Explanation:

Given the concentration of the solution $Ba(OH)_{2}=0.050M$

The dissociation reaction of $Ba(OH)_{2}$ can be written as

                          $Ba(OH)_{2} \rightarrow Ba^{2+} +2OH^{-}$

From the reaction, 1 $Ba(OH)_{2}$ molecule gives rise 2$OH^{-}$ molecule,

i.e., the concentration of molecules is

                          $Ba(OH)_{2}:(OH)^{-}$ is $1:2$

therefore the concentration of $OH^{-}$ molecule is

                                   $0.050\times 2=0.1M$

Using the self ionization of water, where the product of the molarity of hydronium and hydroxide ion,  

                       $[H_{3} O^{+} ][OH^{-} ]=1\times10^{-14}$

concentration of $H_{3} O^{+}$ is,

                               $c[H_{3} O^{+} ]=\frac{1\times10^{-14}}{c[OH^{-} ]}$

                                             $=\frac{1\times10^{-14}}{0.1}=10^{-13}M$

Therefore, concentration of  $H_{3} O^{+}$ in  $Ba(OH)_{2}$ solution is $10^{-13}M$.