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# What is produced when a base is dissolved in water?

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?

#### Explanation

Explain in detail...

#### Explanation:

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anor277 Share
Jun 19, 2018

$\text{Hydroxide ions...?}$

#### Explanation:

We know that water undergoes a measurable autoprotolysis...i.e.

$2 {H}_{2} O \left(l\right) r i g h t \le f t h a r p \infty n s {H}_{3} {O}^{+} + H {O}^{-}$

For which under standard conditions of $298 \cdot K$, and $100 \cdot k P a$..

${K}_{w} = {10}^{-} 14 \equiv \left[H {O}^{-}\right] \left[{H}_{3} {O}^{+}\right]$

And using standard logarithmic terms... $14 = p H + p O H$, where $p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right]$, and $p O H = - {\log}_{10} \left[H {O}^{-}\right]$

Of course, when we add a soluble hydroxide to solution, for instance $K O H \left(s\right)$ or $N a O H \left(s\right)$ WE INCREASE $\left[H {O}^{-}\right]$...and increase $p H$ BUT NUMERICALLY DECREASE $p O H$...

We COULD add a base such as ammonia, which undergoes an equilibrium reaction with water...

$N {H}_{3} \left(a q\right) + {H}_{2} O \left(l\right) r i g h t \le f t h a r p \infty n s N {H}_{4}^{+} + H {O}^{-}$

Here, since the reaction is NOT quantitative, and governed by an equilibrium constant, ${K}_{b}$, significant concentration of ammonia will remain in solution....$p H$ will be elevated above $7$ due to the presence of SOME $H {O}^{-}$ ions....

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