What is the pOH concentration of a .0768 M NaOH solution?

1.2

Explanation:

pOH is the negative logarithm of hydroxide ion concentration in the solution.

Thus, pOH is expressed as

pH = - log(OH-)

NaOH(aq) ---> Na(aq) + OH-(aq)

The alkali ionizes to give hydroxide ion OH- as the only negative ion. Hence, the concentration of OH- is also 0.0768

So, pOH = - log(0 .0768)

pH = -(-1.15)

pH = 1.15 (since the two minus signs will cancel out)

Round the pOH value to the nearest tenth as 1.2

Thus, the pOH concentration of 0 .0768 NaOH solution is 1.2 (highly concentrated)

the potential energy of the reactants is 200j.

from the energy diagram, the energy of the product formed is 350j; this means that, this reaction is an endothermic reaction, because it absorbs energy from its environment.