Question

How much ammonium chloride do we need to add to a `20*mL` volume of `NH_3(aq)` at `0.50*mol*L^-1` concentration, to maintain a `pH=9.0`?

Answer 1

We need (i) approx. `270*mg`, and (ii) we need to use the buffer equation: `pH=pK_a+log_10(([NH_4^+])/([NH_3]))`.

I do not think the given answer is kosher.

Explanation:

Now we know that `pK_a+pK_b=14` for an aqueous solution under standard conditions, `K_b=2xx10^-5`, and thus `pK_b=4.70`, and `pK_a=9.30`

And we substitute the `pK_a` value back into the original equation, with the desired `pH`.

And thus,

`9=9.30+log_10(([NH_4^+])/([NH_3]))`

And thus `-0.30=log_10(([NH_4^+])/([NH_3]))`

`10^(-0.3)=([NH_4^+])/([NH_3])`

And we know that `[NH_3]=0.50*mol*L^-1`.

So `[NH_4^+]=10^(-0.3)xx0.50*mol*L^-1=0.25*mol*L^-1`

Now `"concentration"` `=` `("Moles of solute")/("Volume of solution")`, but the volume of solution was SPECIFIED to be `20xx10^-3L`.

So `"moles of solute"=20xx10^-3cancelLxx0.25*mol*cancel(L^-1)`

`=5.01xx10^-3*mol` of `NH_4Cl`; and this represents a mass of

`5.01xx10^-3*molxx53.49*g*mol^-1=0.268*g`.

So a final check,

`[NH_4^+]=((0.268*g)/(53.49*g*mol^-1))/(20xx10^-3*L)=0.251*mol*L^-1`.

`[NH_3]=0.50*mol*L^-1`.

Resubstituting into the buffer equation:

`pH=pK_a+log_10(((0.251*mol*L^-1))/((0.50*mol*L^-1)))`

`=9.30+(-0.3)`

`=9.0` AS REQUIRED....................