# Concentration of Hydrogen Ions

There are 4 different solutions ($$\ce{NH3}$$, $$\ce{HCl}$$, $$\ce{NaOH}$$, $$\ce{CH3COOH}$$), their concentration is the same $$(c = \pu{1 mol L-1}).$$ Which solution has pH that is higher than 7, but lower than 14?

It is obvious that $$\ce{HCl}$$ and organic acid cannot have such concentration of hydrogen ions. Both ammonia and natrium hydroxide have base properties. $$\ce{NH3}$$ is a weaker base though. I am not sure how to find the concentration of hydroxide in $$\ce{NH3}$$ and $$\ce{NaOH}$$ properly.

Thank you for the explanation.

Since $$\ce{NaOH}$$ is a strong base, the pH can be directly calculated from the analytic concentration (1M):

$$\ce{pOH} = -\log\ce{[OH-]} = -\log(1) = 0$$

$$\ce{pH} = 14 - \ce{pOH} = 14 - 0 = 14$$

For ammonia, you need to consider its reaction in water and the associated $$\ce{K_b}$$ value,

$$\ce{NH3 + H2O -> NH_4+ + OH-}$$

The equilibrium concentrations for an unknown $$x$$ reacted amount (in concentration units) would be

$$\ce{[NH3]} = \ce{[NH3]_0} - x$$

$$\ce{[OH-]} = x$$

$$\ce{[NH4+]} = x$$

$$\ce{K_b = \frac{[NH4+][OH-]}{[NH3]} = \frac{x^2}{[NH3]_0 - x}}$$

By plugging in the proper $$\ce{K_b}$$ value for ammonia and its analytical concentration ($$\ce{[NH3]_0}$$), you can calculate $$x$$, from which

$$\ce{pOH} = -\log(x)$$