How much ammonium chloride do we need to add to a $20 \cdot m L$ $20mL$ volume of $N H_{3} (a q)$ $NH_3(aq)$ at $0.50 \cdot m o l \cdot L^{- 1}$ $0.50mol*L^-1$ concentration, to maintain a $p H = 9.0$ $pH=9.0$ ?

Question

How much ammonium chloride do we need to add to a $20 \cdot m L$ $20mL$ volume of $N H_{3} (a q)$ $NH_3(aq)$ at $0.50 \cdot m o l \cdot L^{- 1}$ $0.50mol*L^-1$ concentration, to maintain a $p H = 9.0$ $pH=9.0$ ?

1 Answer

Impact of this question

2046 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-08T16:40:50

We need (i) approx. $270 \cdot m g$ $270*mg$ , and (ii) we need to use the buffer equation: $p H = p K_{a} + {log}_{10} (\frac{[N H_{4}^{+}]}{[N H_{3}]})$ $pH=pK_a+log_10(([NH_4^+])/([NH_3]))$ .

I do not think the given answer is kosher.

Explanation:

Now we know that $p K_{a} + p K_{b} = 14$ $pK_a+pK_b=14$ for an aqueous solution under standard conditions, $K_{b} = 2 \times 10^{- 5}$ $K_b=2xx10^-5$ , and thus $p K_{b} = 4.70$ $pK_b=4.70$ , and $p K_{a} = 9.30$ $pK_a=9.30$

And we substitute the $p K_{a}$ $pK_a$ value back into the original equation, with the desired $p H$ $pH$ .

And thus,

$9 = 9.30 + {log}_{10} (\frac{[N H_{4}^{+}]}{[N H_{3}]})$ $9=9.30+log_10(([NH_4^+])/([NH_3]))$

And thus $- 0.30 = {log}_{10} (\frac{[N H_{4}^{+}]}{[N H_{3}]})$ $-0.30=log_10(([NH_4^+])/([NH_3]))$

$10^{- 0.3} = \frac{[N H_{4}^{+}]}{[N H_{3}]}$ $10^(-0.3)=([NH_4^+])/([NH_3])$

And we know that $[N H_{3}] = 0.50 \cdot m o l \cdot L^{- 1}$ $[NH_3]=0.50*mol*L^-1$ .

So $[N H_{4}^{+}] = 10^{- 0.3} \times 0.50 \cdot m o l \cdot L^{- 1} = 0.25 \cdot m o l \cdot L^{- 1}$ $[NH_4^+]=10^(-0.3)xx0.50*mol*L^-1=0.25*mol*L^-1$

Now $concentration$ $"concentration"$ $=$ $=$ $\frac{Moles of solute}{Volume of solution}$ $("Moles of solute")/("Volume of solution")$ , but the volume of solution was SPECIFIED to be $20 \times 10^{- 3} L$ $20xx10^-3L$ .

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

Science

Math

Humanities

... and beyond

How much ammonium chloride do we need to add to a $20 \cdot m L$ $20mL$ volume of $N H_{3} (a q)$ $NH_3(aq)$ at $0.50 \cdot m o l \cdot L^{- 1}$ $0.50mol*L^-1$ concentration, to maintain a $p H = 9.0$ $pH=9.0$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How much ammonium chloride do we need to add to a 20*mL20⋅mL20*mL volume of NH_3(aq)NH3(aq)NH_3(aq) at 0.50*mol*L^-10.50⋅mol⋅L−10.50*mol*L^-1 concentration, to maintain a pH=9.0pH=9.0pH=9.0?

1 Answer

Explanation:

Related questions

Impact of this question

How much ammonium chloride do we need to add to a $20 \cdot m L$ $20mL$ volume of $N H_{3} (a q)$ $NH_3(aq)$ at $0.50 \cdot m o l \cdot L^{- 1}$ $0.50mol*L^-1$ concentration, to maintain a $p H = 9.0$ $pH=9.0$ ?