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Khouuu · Jun 10, 2009, 03:15 PM

Calculate the hydronium ion concentration and the pH when 50.0 mL of 0.40 M NH3 is mixed with 50.0 mL of 0.40 M HCl.

Perito · Jun 10, 2009, 08:13 PM

$pH = -log([H^+])$

where $[H^+]$ is the hydrogen ion concentration. Hydronium ion is $H_3O^+$ but for all intents and purposes, the hydrogen ion concentration is the hydronium ion concentration. $[H^+] = [H_3O^+]$

50 mL of 0.40 M NH3: 50 mL x 0.40 mmoles/mL = 20 mmoles NH3

50 mL of 0.40 M HCl: 50 mL x 0.40 mmoles/mL = 20 mmoles HCl

$NH_3 + H_2O \rightleftharpoons NH_4OH \rightleftharpoons NH_4^+ + OH^-$

$HCl \rightleftharpoons H^+ + Cl^-$

$H^+ + OH^- \rightleftharpoons H_2O$

If you follow through the equations, you'll see that you have the same amount of OH- as you do H+. Since you have exactly the same amount, you'll have pure water + [math]NH_4Cl[math] in solution. If you ignore the effect of the salt (risky. With the information provided in this problem, there appears to be no other way to handle it here.), you wind up with simply water. No excess hydroxide, no excess hydrogen ion.

$[H^+] \times [OH^-] = 10^{-14}$

Since $[H^+] = [OH^-]$ then $[H^+] = 10^{-7)$

The pH is, therefore $-log(10^{-7}) = 7$