# Help: Equilibrium 1

$40 mL$, $0.1 M$ ammonia solution is mixed with $20 mL$ $0.1 M HCl$ , what is the $pH$ of mixture ?($pK_{a}$ of $NH_{3}$ solution =$4.74$)

Now, what I did was as follows:-

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

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

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

After obtaining the $\left[ { OH }^{ - } \right]$, I found the moles of ${ OH }^{ - }$ and subtracted it from the moles of ${ H }^{ + }$ and the and found the conc. of ${ H }^{ + }$, 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 4 years, 6 months ago

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

- 4 years, 6 months ago

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.

- 4 years, 6 months ago