Calculate pH of NH3/NH4Cl Buffer System After NaOH Addition

[Mitchtwitchita]

Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system. What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80 mL of the buffer solution?

NH4+ ---> NH3 + H+

0.36 0.30 0
-x +x +x
0.36-x 0.30 + x x

Ka=[NH3][H+]/[NH4+]
5.6 x 10^-10 = (0.30 + x)(x)/(0.36 - x)
5.6 x 10^-10 = 0.30x/0.36, assuming x is negligible
=6.72 x 10^-10

pH=-log[H+]
=-log (6.72 x 10^-10)
=9.17

I don't know what to do next, can anybody please point me in the right direction?

 

[Snazzy]

For the second part, the OH- will react with the NH4+ ion to produce NH3 + H2O, so the number of moles of acid decreases by an equal amount as the number of moles of OH- and the number of moles of base increases by the same amount, so:

[H3O+] = Ka x (number of moles of acid) / (number of moles of conjugate base)

[H30+] = (Ka)((0.36)(0.08) - (0.050)(0.0200)) / ((0.30)(0.08) + (0.050)(0.0200))

So you get a pH of 9.21; barely any change at all.

 

[Mitchtwitchita]

Thanks Snazzy!

 

[AbedeuS]

Whats with the x's in the first part? I thought it would be as simple as:
$$Ka = \frac{[NH_{3}][H^{+}]}{[NH_{4}^{+}]} = 5.6*10^{-10}$$

This isn't me saying my take on the answer btw, I just haven't done buffer stuff in ages and would like to know

 

[Mitchtwitchita]

Hey AbedeuS,

We were taught to use an I.C.E. table inorder to answer these types of questions, where I is the initial molar concentration, C is the change in concentraion and E is the concentration at equilibrium. The x's are used in place of unknown quantities. I'm going to try your way though, because I'm all for simplicity! Cheers.

 

[AbedeuS]

I'm not too sure if my way is correct is the problem though, I'm just going off basic equilibria principles so:

$$K_{a} = \frac{[NH_{3}][H^{+}]}{[NH_{4}^{+}]}$$

Substitute and rearrange:

$$[H^{+}] = \frac{5.6*10^{-10} * 0.36}{0.3} = 6.72*10^{-10}$$

Convert concentration to pH:

$$-log(6.72*10^{-10}) = 9.1726$$

Part 2,

Since your adding a strong base (NaOH) it can be presumed that it deprotinates an amount of ammonium equal to its own number of moles leaving an equivalent amount of ammonia which will be added to the origional amount. Remember however that Ka is working with concentrations, so converting to moles to add and take away quantities does require division by the new volume to convert back into concentration using the $$n = cv$$ equation Therefore:

$$[NH_{4}^{+}]_{new} = \frac{([NH_{4}^{+}]_{old}*V_{NH_{4}^{+}}) - ([OH-]*V_{NaOH})}{V_{new}}$$

And:

$$[NH_{3}]_{new} = \frac{([NH_{3}]_{old}*V_{NH_{3}})+([OH^{-}]*V_{NaOH})}{V_{new}}$$

But since the acid equilibrium equation has $$NH_{4}^{+}$$ on the bottom and $$NH_{3}$$ on the top, the division by the new volume to give the new concentration is unnecessary and cancels out leading too:

$$6.72*10^{-10} = \frac{[NH_{3}][H^{+}]}{[NH_{4}^{+}]} = \frac{[0.3*0.08 + 0.050*0.02][H^{+}]}{[0.36*0.08 - 0.05*0.02]}}$$

Rearranging:

$$\frac{5.6*10^{-10}*(0.36*0.08-0.05*0.02)}{0.3*0.08+0.05*0.02} = [H^{+}] = 6.2272*10^{-10}$$

Taking logs:

$$pH = -log(6.2272*10^{-10}) = 9.206$$

This is just the long winded way of what Snazzy did, but buffer solutions tend to pop up a lot and I wanted a referral type answer.

EDIT: Subbed in wrong Ka, leading to a lower pH then expected.

 

[Mitchtwitchita]

Thanks AbedeuS,

Now I know where I was going wrong. I saw it in snazzy's response also, but now I know precicesly where and why. Thanks for your help guys!

 

[AbedeuS]

I think the X's in your table are used because the concentrations I have used for the maths just uses the question concentrations, the reality I presume involves solving two Ka equations, one for pure NH4 and another for when the pure NH4 is put with NH3, ill have a look later.