50 mL of 0.1 M NH3 + 50 mL of 0.1 M NH4NO3.
(NH3) = 0.1 x (50/100) = 0.05 M.
(NH4NO3) = 0.1 x (50/100) = 0.05 M in the buffer.
1)So we start with 10 mL 0.05 M buffer. We dilute NH3 and NH4NO3 with 6 mL water.
The new concn (NH3) = 0.05 x (10/16) = ??
The new concn (NH4NO3) = 0.05 x (10/16) = ??
pH = pKa + log [(base)/(acid)]
Solve for pH. (The bottom line to dilution with water only is that the pH doesn't change but you go through a lot of concn changes to show that.)
10 mL buffer gives me how many moles NH3? That is M x L = 0.05 x 0.01 L = 0.0005 moles NH3.
How many moles NH4NO3. 0.05 x 0.01 L = 0.0005 moles NH4NO3.
How much HCl is being added? That is M x L = 0.1 M x 0.001 L = 0.0001 moles HCl.
How much of the NH3 is used? 0.0001 moles HCl will react with 0.0005 moles NH3 to form 0.0001 moles NH4NO3 (to make a total of 0.0001+0.0005 = 0.0006 moles NH4NO3) and it will leave 0.0005 - 0.0001 = 0.0004 moles NH3).
pH = pKa + log(base/acid)
pH = 9.26 + log(0.0004/0.0006) = 9.08
Check my work. It's so easy to omit a zero. It is much easier to work with millimoles and not mols. M x mL = mmoles and M = mmoles/mL. For example 10 mL x 0.05 M = 0.5 mmol NH3. We add 1 mL x 0.1 M HCl = 0.1 mmole. So we form 0.1 + 0.5 = 0.6 mmole NH4NO3 and we leave 0.5-0.1 = 0.4 mmole NH3. Now, I ignored the water which was added. If you want to put in concn, then the new NH3 concn = 0.4 mmole/16 mL = ?? M and the new NH4NO3 concn = 0.6 mmole/16 mL = ?? M. You can see the 16 mL cancels.The others are done essentially the same way.