Question

What is the pOH of 2.15 M Ba(OH)2 solution? Is the solution neutral, acidic, or basic?

Answer 1

Explanation:

Given

2.15⋅mol⋅L^−1

barium hydroxide,

[HO−]=4.30⋅mol⋅L^−1

pOH =−log10[HO−]=−log10{4.30} =−0.63

The solution is strongly basic.

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Answer 2

Answer:

The pOH of $2.15M$ $Ba(OH)_{2}$ solution measured is $-0.63$.

The solution is very strongly basic.

Explanation:

Given,

The concentration of $[OH^{-}]$ = $2.15M$

The pOH of $2.15M$ $Ba(OH)_{2}$ solution =?

As we know,

• $Ba(OH)_{2}$ is a strong base.

Thus, it completely dissociates into ions as shown below:

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

As given,

• $1[OH^{-}]$ = $2.15M$
• $2[OH^{-}]$ = $2\times 2.15$ = $4.3M$

Now,

• pOH = $-log[OH^{-}]$
• pOH = $-log[4.3]$
• pOH = $-log[\frac{43}{10}]$
• pOH = $log \frac{10}{43}$                      
• pOH = $log 10 - log 43$             ( $log\frac{a}{b} = log a - logb$ )
• pOH = $1- 1.63$                     ( $log 10 = 1, log 43 = 1.63$ )
• pOH = $-0.63$

Now, $pH + pOH = 14$

• $pH + -0.63 = 14$
• $pH = 14.63$

This means that the solution is very strongly basic.