\(40 mL\), \(0.1 M\) 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

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestYour 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.

Log in to reply

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

Log in to reply