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I am slightly confused about one certain task.

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.

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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

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

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