# Question f26bd

Jul 24, 2016

$p H = 12.40$

#### Explanation:

Assuming you meant $N a O H$, to solve that, we must first get the molarity of the solution, i.e.: the concentration in moles per liter.

0.1%# is the same as saying we have $0.1 g$ per $100 m L$ so, we have

$C = \frac{0.1 g}{100 m L} = \frac{1 g}{1000 m L}$

As $1000 m L = 1 L$ we can say that $C = 1 g \cdot m {L}^{- 1}$, from there we divide by the molar mass to get the molarity

$C = \frac{1 g}{1 L} \cdot \frac{1 m o l}{40 g} = 0.025 m o l \cdot {L}^{- 1}$

That's the concentration of $N a O H$, but since the concentration of liberated $O {H}^{-}$ ions is the same as the concentration of the compound as a whole (since we have a strong base and a high enough concentration) we can say that

$p O H = - \log \left(C\right)$

And that

$p H = 14 - p O H$

From the water equilibrium constant ${K}_{w} = \left[{H}^{+}\right] \left[O {H}^{-}\right] = {10}^{- 14}$

Or

$p H = 14 - \left(- \log \left(C\right)\right) = 14 + \log \left(C\right)$

Switching the values in, we have

$p H = 14 + \left(- 1.60\right) = 14 - 1.60 = 12.40$

We can leave it at that, although proper chemistry would require you to aproximate that to the approximate number of sig figs, in this case, 1. So the answer would be $1 \cdot {10}^{1}$, but that's clunky, unnecessary and most likely not what your teacher wanted.