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I am slightly confused about one certain task. There are 4 different solutions (NH3, HCl, NaOH, CH3COOH), their concentration is the same ( c=1 mol/l). The question is which solution has pH that is higher than 7, but lower than 14. It is obvious that HCl and organic acid cannot have such concentration of hydrogen ions. Both ammonia and natrium hydroxide have base properties. NH3 is a weaker base though. I am not sure how to find the concentration of hydroxyde in NH3 and NaoH properly. Thank you for the explanation.

Answer 1

I am slightly confused about one certain task. There are 4 different solutions (NH3, HCl, NaOH, CH3COOH), their concentration is the same ( $c = 1 \, \text{mol/l}$ ). The question is which solution has pH that is higher than 7, but lower than 14. It is obvious that HCl and organic acid cannot have such concentration of hydrogen ions. Both ammonia and natrium hydroxide have base properties. NH3 is a weaker base though. I am not sure how to find the concentration of hydroxyde in NH3 and NaOH properly. Thank you for the explanation.

pH 1M NaOH

**Solution**

$$\mathrm{c_{NaOH}} = 1 \, \mathrm{M}$$ $$\mathrm{NaOH} \rightleftharpoons \mathrm{Na}^{+} + \mathrm{OH}^{-}$$ $$[\mathrm{OH}^{-}] = 1 \, \mathrm{M}$$ $$\mathrm{pOH} = -\log [\mathrm{OH}^{-}]$$ $$\mathrm{pOH} = -\log [1]$$ $$\log 1 = 0$$ $$\mathrm{pOH} = 0$$ $$\mathrm{pH} = 14 - \mathrm{pOH}$$ $$\mathrm{pH} = 14 - 0 = 14$$

**Answer:** $\mathrm{pH} = 14$

pH 1M ammonia aqueous solution ($\mathrm{NH_3 \cdot H_2O}$)

**Solution:**

$$\mathrm{NH_3 \cdot H_2O} \rightleftharpoons \mathrm{NH_4^+} + \mathrm{OH}^-$$ $$K_b = \frac{[\mathrm{NH_4^+}][\mathrm{OH}^-]}{[\mathrm{NH_3 \cdot H_2O}]} = 1.8 \cdot 10^{-5}$$ $$[\mathrm{NH_4^+}] = [\mathrm{OH}^-]$$ $$[\mathrm{NH_3 \cdot H_2O}] \approx c_{\mathrm{NH_3 \cdot H_2O}}$$ $$K = \frac{[\mathrm{OH}]^2}{[\mathrm{NH_3 \cdot H_2O}]}$$ $$[\mathrm{OH}^{-}] = \sqrt{K_b \cdot C_{\mathrm{NH_3 \cdot H_2O}}} = \sqrt{1.8 \times 10^{-5} \cdot 1} = 0.0042426$$ $$\mathrm{pOH} = -\log [\mathrm{OH}^{-}] = -\log [0.0042426] = -2.37$$ $$\mathrm{pH} = 14 - 2.37 = 11.63$$

**Answer:** $\mathrm{pH} = 11.63$

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