I cannot figure the grams out, i answered parts 2 and 3 correctly. I keep gettin

Question

I cannot figure the grams out, i answered parts 2 and 3 correctly. I keep getting 1.88g for the grams and it is incorrect. I use the determined concentration of the Henderson Hasselbach equation for acetic acid. I then take the molarity of acetic acid and multiply by 500mL or 0.5L to get moles. I take the moles and multiply by molecular weight and get 1.88g of Acetic acid. It is still incorrect

You are asked to prepare 500. mL of a 0.200 M acetate buffer at pH 5.10 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. 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. What volume of 3.00 M NaOH must you add to the acetic acid to achieve a buffer with a pH of 5.10 at a final volume of 500 mL? (Ignore activity coefficients.) 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

Given : [CH3COO-] = 0.200 M, volume = 500 mL = 0.500 L

pH = 5.0

We use Henderson hasselbalch equation

pH = pka + log ( [CH3COO-] / [CH3COOH] )

Lets find concentration of acetic acid first.

5.0 = 4.76 + log ( 0.200/ [CH3COOH] )

We solve for concentration of acetic acid

5.0- 4.76 = log ( 0.200/ [CH3COOH] )

[CH3COOH]= 0.115 M

We know number of moles = volume in L x molarity

Therefore number of moles of acetic acid

= 0.500 L x 0.115 M

=0.0575 mol

Mass of acid = number of moles x molar mass

= 0.0575 mol x 60.05 g/mol

=3.45 g

Mass of acid required = 3.45 g a

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