Question

Determine the poh of a 0.348 m ba(oh)2 solution at 25°c.

Answer 1

Given:

Concentration of Ba(OH)2 = 0.348 M

To determine:

pOH of the above solution

Explanation:

Based on the stoichiometry-

1 mole of Ba(OH)2 is composed of 1 mole of Ba2+ ion and 2 moles of OH- ion

Therefore, concentration of OH- ion = 2*0.348 = 0.696 M

pOH = -log[OH-] = - log[0.696] = 0.157

Ans: pOH of 0.348M Ba(OH)2 is 0.157

Answer 2

0.157

Further explanation

Given:

0.348 M of Ba(OH)₂ solution at 25⁰C

Question:

Determine the pOH of Ba(OH)₂.

The Process:

Ba(OH)₂ represents a strong base. The characteristic of a strong base is it has no ionization constant, Kb or the degree of ionization, α. Ba(OH)₂ is fully ionized in water.

The concentration of OH⁻ ions will determine the pOH value and pH by looking at the number of OH⁻ ions or sometimes referred to as base valence.

Let us prepare the concentration of OH⁻ ions.

$\boxed{ \ Ba(OH)_2_{(aq)} \rightarrow Ba^{2+}_{(aq)} + 2OH^-_{(aq)} \ }$

$\boxed{ \ [OH^-] = \frac{2}{1} \times [Ba(OH)_2] \ }$

$\boxed{ \ [OH^-] = 2 \times 0.348 \ M \ }$

∴ $\boxed{ \ [OH^-] = 0.696 \ M \ }$

After obtaining the concentration of OH⁻ ions, we can immediately determine the pOH value.

$\boxed{ \ pOH = -log[OH^-] \ }$

$\boxed{ \ pOH = -log \ 0.696 \ }$

Thus, we get a pOH of 0.157.

_ _ _ _ _ _ _ _ _ _

What is the pH value of the 0.348 M Ba(OH)₂ solution? Let us continue the rest of the calculation.

Recall that $\boxed{ \ pH + pOH = 14 \ }$

$\boxed{ \ pH = 14 - pOH \ }$

$\boxed{ \ pH = 14 - 0.157 \ }$

∴ $\boxed{ \ pH = 13.843 \ }$

A pH value closes up to 14 indicates that the solution represents a very strong base. In the same way, if the pH value is very close to one, it means the solution is a very strong acid.

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