Buffers DATA TABLE Buffer A Buffer B Mass of NaC2H302 used to prepare buffer (g)

Question

Buffers DATA TABLE Buffer A Buffer B Mass of NaC2H302 used to prepare buffer (g) Volume of buffer prepared (mL) Molar concentration of HC2H302 in buffer (M) Initial pH of buffer Volume of 0.5 M NaOH to raise pH by 2 units 100.0 100.0 0.1 1.0 (mL) Volume of 0.5 M HCI to lower pH by 2 units (mL) Volume of 0.5 M NaOH at equivalence point (mL) DATA ANALYSIS 1. Write reaction equations to explain how your acetic acid-acetate buffer reacts with an acid and reacts with a base 2. Buffer capacity has a rather loose definition, yet it is an important property of buffers. A commonly seen definition of buffer capacity is: "The amount of H or OH that can be neutralized before the pH changes to a significant degree." Use your data to determine the buffer capacity of Buffer A and Buffer B buelten capacity or the ler Anand Blufcer s.ican dese your data to destermine the 3. Say, for example, that you had prepared a Buffer C, in which you mixed 8.203 g of sodium acetate, NaC2H O2, with 100.0 mL of 1.0 M acetic acid. a. What would be the initial pH of Buffer C? b. If you add 5.0 mL of 0.5 M NaOH solution to 20.0 mL each of Buffer B and Buffer C which buffer's pH would change less? Explain. 10-4

Explanation / Answer

acidic buffer buffer A

No of mol of acid = 0.1*0.1   = 0.01 mol

No of mol of salt = 0.147/83 = 0.00166 mol

pka of acid = -log ka = -log (1.8*10^(-5) = 4.74

pH = pka + log(0.00166/(0.01-0.00166))

= 4.74 + log(0.00166/(0.01-0.00166))

= 4.03

pH = 4.03+2 = 6.03

pH = pka + log(0.00166/(0.01-0.00166))

6.03 = 4.74 + log((0.00166+x)/(0.00834-x))

x = 0.0078 mol

Volume of NaOH added = 0.0078/0.5 = 0.0156 L

= 15.6 ml

2.03 = 4.74 + log((0.00166-x)/(0.00834+x))

x = 0.00164 mol

Volume of HCl added = 0.00164/0.5 = 0.00328 L

= 3.28 ml
at equivalence point

No of mol of aceticacid = No of mol of NaOH

No of mol of aceticacid = 0.00834 mol

so that

No of mol of salt = 0.00834+0.00166 = 0.01

pH of salt = 7-1/2(pka+logC)

            = 7+1/2(4.74+log 0.01)

         = 8.37

buffer B

No of mol of acid = 1*0.1    = 0.1 mol

No of mol of salt = 1.476/83 = 0.0177 mol

pka of acid = -log ka = -log (1.8*10^(-5) = 4.74

pH = pka + log(0.00166/(0.1-0.00166))

= 4.74 + log(0.00166/(0.1-0.00166))

   = 2.97

similar to the above do for the buffer B

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