Help: Equilibrium 1

40mL40 mL, 0.1M0.1 M ammonia solution is mixed with 20mL20 mL 0.1MHCl0.1 M HCl , what is the pHpH of mixture ?(pKapK_{a} of NH3NH_{3} solution =4.744.74)

Now, what I did was as follows:-

First of all I calculated the [OH][OH^{-}] obtained from the ammonia solution. Which i did as follows NH3+H2ONH4++OHc00ccαcαcαNH_{ 3 }\quad \quad \quad \quad +\quad \quad \quad \quad H_{ 2 }O\quad \quad \quad \quad \leftrightharpoons \quad \quad \quad \quad NH_{ 4 }^{ + }\quad \quad \quad \quad +\quad \quad \quad \quad { OH }^{ - }\\ \quad c\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 0\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 0\\ c-c\alpha \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad c\alpha \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad c\alpha

From above we can conclude that [OH]=Kbc\left[ { OH }^{ - } \right] =\sqrt { { K }_{ b }c } (We get Kb{ K }_{ b } from the relation Kb+Kc=14{ K }_{ b }+{ K }_{ c }=14)

The explanation is getting bigger so i will cut to the chase.

After obtaining the [OH]\left[ { OH }^{ - } \right], I found the moles of OH{ OH }^{ - } and subtracted it from the moles of H+{ H }^{ + } and the and found the conc. of H+{ H }^{ + }, then the -ve logarithm. But the book gives a different answer from what i got. So if anything that i did wrong or left somewhere could you point it out

Note by Akhilesh Prasad
5 years, 4 months ago

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@Rishabh Cool. If you can also look into this one please.

Akhilesh Prasad - 5 years, 4 months ago

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Your equilibrium condition is not considering HCl. The equilibrium shifts much more to the right to produce hydroxide ions in presence of the acidic HCl. This solution acts as a buffer solution.

Ameya Daigavane - 5 years, 4 months ago

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