Calculate pH of NH3/NH4Cl Buffer System After NaOH Addition

In summary: I really need to get to a chem class soon. School is always so busy. Cheers for now.In summary, the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system is 9.17. After the addition of 20.0 mL of 0.050 M NaOH to 80 mL of the buffer solution, the pH remains relatively unchanged at 9.21 due to the buffering capacity of the system. This is because the addition of the strong base only slightly shifts the equilibrium between NH4+ and NH3, resulting in a minimal change in the concentration of H+.
  • #1
Mitchtwitchita
190
0
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?
 
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  • #2
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.
 
  • #3
Thanks Snazzy!
 
  • #4
Whats with the x's in the first part? I thought it would be as simple as:
[tex]Ka = \frac{[NH_{3}][H^{+}]}{[NH_{4}^{+}]} = 5.6*10^{-10} [/tex]

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 o_O
 
  • #5
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.
 
  • #6
I'm not too sure if my way is correct is the problem though, I'm just going off basic equilibria principles so:

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

Substitute and rearrange:

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

Convert concentration to pH:

[tex] -log(6.72*10^{-10}) = 9.1726 [/tex]

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 [tex] n = cv [/tex] equation Therefore:

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

And:

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

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

[tex] 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]}} [/tex]

Rearranging:

[tex] \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}[/tex]

Taking logs:

[tex] pH = -log(6.2272*10^{-10}) = 9.206 [/tex]

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.
 
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  • #7
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!
 
  • #8
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.
 

FAQ: Calculate pH of NH3/NH4Cl Buffer System After NaOH Addition

1. How do I calculate the pH of an NH3/NH4Cl buffer system after adding NaOH?

To calculate the pH of an NH3/NH4Cl buffer system after adding NaOH, you will need to use the Henderson-Hasselbalch equation. This equation relates the pH of a buffer solution to the pKa of the weak acid and the ratio of the weak acid and its conjugate base. The equation is pH = pKa + log([A-]/[HA]), where [A-] is the concentration of the weak base (NH3) and [HA] is the concentration of the conjugate acid (NH4+). You will also need to consider the concentration of NaOH added and the initial concentrations of NH3 and NH4Cl in the buffer system.

2. What is the pKa of NH4Cl?

The pKa of NH4Cl is not a fixed value, as the pKa of a substance can vary depending on the solvent and temperature. However, the pKa of NH4Cl in water at 25°C is approximately 9.25. It is important to use the correct pKa value for the Henderson-Hasselbalch equation to accurately calculate the pH of the buffer system.

3. How does adding NaOH affect the pH of an NH3/NH4Cl buffer system?

Adding NaOH to an NH3/NH4Cl buffer system will increase the pH of the solution. This is because NaOH is a strong base and will react with the weak acid, NH4+, to form the conjugate base, NH3. This will shift the equilibrium towards more NH3, resulting in a higher pH.

4. What is the purpose of using a buffer system in this experiment?

A buffer system is used to maintain a stable pH in a solution. In this experiment, the NH3/NH4Cl buffer system is used to resist changes in pH when small amounts of acid or base are added. This is important in many biochemical and chemical processes, as a stable pH is necessary for optimal reaction rates and to prevent damage to biological molecules.

5. How can I adjust the pH of the buffer system if it is not at the desired level?

If the pH of the buffer system is not at the desired level, you can adjust it by adding either a strong acid or a strong base. This will change the ratio of the weak acid and its conjugate base, and therefore change the pH. It is important to use small increments of the acid or base and monitor the pH carefully to avoid over-correction.

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