Question

You are asked to bring the pH of 0.500L of 0.550 M NH4Cl to 7.00. Which of the following solutions would you use: 12.0 M HCl or 12.0 M NH3? How many drops (1 drop 0.05 mL ) of this solution would you use?
The answer to part one is 12.0 M NH3 but how do you answer the second part?

Thank you

Answer 1

Ok, so you've figured out that ammonium chloride, once dissolved, creates an acidic solution. So now you need to find out how much it actually changes the pH. To do this, let's look at the following reaction...

\[NH_4^+ (aq)+H_2O(l) \rightleftharpoons NH_3(aq)+H_3O^+(aq)\]What we are doing when we add the ammonia is that we are creating a buffer solution. We can use the Henderson-Hasselbalch equation for this...\[pH=pKa+\log(\frac{[NH_3]}{[NH_4^+]})=pKa+\log(\frac{n_{NH_3}}{n_{NH_4^+}})\]
In this case, we know the pH of the solution, we can calculate pKa easily, and we know the concentration of ammotables), and we can calculate the pKa of ammonium. \[K_b(NH_3)=1.8 \times 10^{-5} \]From this value, we can calculate the K_a of ammonia's conjugate acid, which is ammonium... \[K_a(NH_4^nium (in terms of the concentration of ammonia).

So first, let's find the Kb for ammonia (you generally won't find the Ka for ammonium in Ka +)=\frac{1.0 \times {10^{-14}}}{1.8 \times {10^{-5}}}=5.6 \times {10^{-10}}\]\[pKa(NH_4^+)=9.26\]Now, let's take a look back at the Henderson-Hasselbalch equation (the one in terms of moles)...\[pH=pKa+\log(\frac{NH_3}{NH_4^+})\]We know that the number of moles of ammonium is going to stay constant, so we only need to worry about the moles of ammonia. This value can be modeled by...\[[NH_3]=\frac{n_{NH_3}}{V_{sol'n}} \rightarrow n_{NH_3}=([NH_3])(V_{sol'n})\]So we essentially need to find out how many moles of ammonia are in each drop of it (0.05mL)...\[n_{NH_3}=(12.0M)(x*0.00005L)=0.0006x\]Where x is the number of drops of ammonia that has been added. Let's go ahea

Answer 2

Sorry, my browser window froze and I thought it crashed, so I hit "post" real quick. Let me continue...

Let's go ahead and calculate the number of moles of ammonium...\[[NH_4^+]=\frac{n_{NH_4^+}}{V_{sol'n}} \rightarrow n_{NH_4^+}=([NH_4^+])(V_{sol'n})\]\[n_{NH_4^+}=(0.550M)(0.500L)=0.275mol\]And let's substitute back in some numbers into Henderson-Hasselbalch...\[7.00=9.26+\log(\frac{0.0006x}{0.275})\]\[-2.26=\log(\frac{0.0006x}{0.275})\]\[0.00556=\frac{0.0006x}{0.275} \rightarrow x = 2.5\ drops\]

Answer 3

Thank you but for some reason, the online program says this is wrong answer,

Answer 4

Perhaps try whole numbers close to 2.5, as it may not always be possible to add half a drop (depending on your mechanism). So try 2 and 3 drops, I guess.