# A 0.45*g mass of HCl is dissolved in water to give a 300*mL volume of solution. What are pOH, and [HO^-] for this solution?

Mar 22, 2017

We interrogate the equilibrium:

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

#### Explanation:

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

This equilibrium lies almost entirely to the right, and we assume the solution is stoichiometric in ${H}_{3} {O}^{+}$.

And thus $\left[H C l\right] = \frac{\frac{0.45 \cdot g}{36.46 \cdot g \cdot m o {l}^{-} 1}}{300.0 \times {10}^{-} 3 L} = 0.041 \cdot m o l \cdot {L}^{-} 1$.

Because dissociation is QUANTITATIVE, $\left[{H}_{3} {O}^{+}\right] = \left[H C l\right] = 0.041 \cdot m o l \cdot {L}^{-} 1$.

And $p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right] = 1.39$

$p O H = 14 - p H = 12.61$, and thus $\left[H {O}^{-}\right] = {10}^{- 12.61} = \text{something small}$