posted by Mary on .
How do you calculate the pH when the following substances are added to a buffer.(all of the solutions are at .10M conc.)
The buffer: 50mL NH3 + 50mL NH4NO3
1) 10mL buffer + 6mL water
2)10mL buffer + 5mL water +1mL HCl
3)10mL buffer + 6mL HCl
4)10mL buffer + 5mL water + 1mL NaOH
I'm really confused on how the equations are set up for these types of problems (buffer solns).I'm posting this for the third time and really need some help..
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.