I just need the answer to #2, I just put the other problems for the information.
ID: 501006 • Letter: I
Question
I just need the answer to #2, I just put the other problems for the information.
You are asked to prepare 500. mL of a 0.200 M acetate buffer at pH 4.90 using only pure acetic acid (MW = 60.05 g/mol, pK_a = 4.76), 3.00 M NaOH, and water. Answer the following questions regarding the preparation of the buffer. 1. How many grams of acetic acid will you need to prepare the 500 mL buffer? Note that the given concentration of acetate refers to the concentration of all acetate species in solution. 2. What volume of 3.00 M NaOH must you add to the acetic acid to achieve a buffer with a pH of 4.90 at a final volume of 500 mL? (lgnore activity coefficients.) 3. Pour the beaker contents into a 500 mL volumetric flask. Rinse the beaker several times and add the rinses to the flask. Swirl the flask to mix the solution. Add water to the mark to make the volume 500 mL. Stopper the flask and invert how many times to ensure complete mixing?Explanation / Answer
Ac- = conjugate acid of Acetic acid
HAc=Acetic acid
The reaction between acetic acid and NaOH is as followed:
HAc + NaOH -------------------> NaAc + H2O
1 mole of NaOH is needed to make 1 mole of Ac- from HAc
pH = pka + log [Ac-/HAc]
4.90 = 4.76 + log[Ac-/HAc]
log[Ac-/HAc] = 0.14
[ Ac-/HAc] = 100.14
[ Ac-/HAc] = 1.38
500 mL=0.5 L
Total moles in buffer=0.5 L x 0.20M=0.1 moles
HAc+Ac^-=0.1 moles
and
[ Ac-/HAc] = 1.38
1.38HAc = Ac-
You have two equations:
1.) HAc + Ac-=0.1 moles
2.) 1.38HAc = Ac-
Substitute 2 into 1 and solve for HAc:
HAc + (1.38HAc)=0.1 moles
2.38HAc=0.1 moles
HAc=0.1/2.38 moles
HAc=0.042 moles
So, you need 0.1 - 0.042 =0.058 moles of Ac-
NaOH is a strong base and will deprotanate HAc to give Ac-, so 0.058 moles of NaOH are needed.
Solving for volume to obtain volume of NaOH solution needed:
moles/volume=Molarity
volume = moles / Molarity
So, volume of NaOH = 0.058/3
= 0.02 L = 20 mL
20 mL of 3 M NaOH solution