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

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Answer 1 · 2017-03-08T16:40:50

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.

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....................

How much ammonium chloride do we need to add to a 20*mL20*mL volume of NH_3(aq)NH_3(aq) at 0.50*mol*L^-10.50*mol*L^-1 concentration, to maintain a pH=9.0pH=9.0?