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In: Chemistry

Three buffers are made by combining a 1M solution of acetic acid with a 1M solution...

Three buffers are made by combining a 1M solution of acetic acid with a 1M solution of sodium acetate in the ratios shown below.

1M acetic acid 1M sodium acetate

Buffer 1: 10mL 90mL

Buffer 2: 50mL 50mL

Buffer 3: 90mL 10mL

If the pKa of acetic acid if 4,75, what is the pH of buffers 1-3?

For this I got Buffer 1 = 5.70 Buffer 2 = 4.75 and Buffer 3 = 3.97

This next part is where I am getting stuck.

What is the pH of Buffers 1-3 when 10mL of 2M NaOh is added?

Thank you for your help in advance!

Expert Solution

since pH= pKa+ log [acetate ion/acetic acid]

moles of acetic acid= molarity* Volume in L= 1*10/1000 , moles of sodium acetate= 1*90/1000

after mixing , the volume= 10+90=100 ml

concentration of acetic acid after mixing = 1*10/1000/(100/1000)

concentration of sodium acetate = 1*90/100/(100/1000)

hence pH= 4.75+ log (90/10)= 5.70, the equation suggests that for calculating pH it is enough to take the ratio of volume of sodium acetate and acetic acid

hence for the second case, pH= 4.75+log (50/50)= 4.75, for the third case, pH= 4.75+log (10/90)= 3.79

3. when NaOH is added, the reaction is CH3COOH+NaOH--------->CH3COONa+ H2O, more of soduium acetate is formed at the expense of acetic acid .

1 mole of acetic acid reacts with 1 mole of sodium hydroxide.

moles of sodium hydroxide= molarity* volume in L= 2*10/1000= 0.02

1. Case 1: mole of acetic acid = 1*10/1000=0.01, mole of sodium acetate= 1*90/1000=0.09

here sodium hydorixie is excess reactant. All the acetic acid is consumed, since sodium hydroxide is strong base, remaining sodium hydorxide= 0.02-0.01=0.01, volume of solution after mixing= 110ml= 0.11L. Concentration of NaOH= 0.01/0.11= 0.091

poH= -log (OH-)= -log (0.091)= 1.04, pH= 14-1.04= 12.96

2. Case 2: mole of acetic acid 1*50/1000= 0.05, mole of sodium hydroxide is limiting, hence moles of sodium acetate formed due to reaction =0.02, total of sodium acetate= 0.03( remaining)+ 0.05= 0.08

volume after mixing = 110ml=0.11L, concentrations after mixing : sodium acetate= 0.08/0.11 and that of acetic acid =0.03/0.11

pH= 4.75+log (0.08/0.03)= 5.2

3. moles of acetic acid in 90ml= 90*1/1000 =0.09, moles of sodium hydroxide =0.02 , excess is acetic acid , moles of sodium acetate= 0.02+0.01=0.03, mole of acetic acid remaining = 0.09-0.02=0.07

concentrations after mixing = 0.03/0.11 and that of acetic acid =0.07/0.11

pH= 4.75+ log (0.03/0.07)= 4.4

queen_honey_blossom answered 2 years ago

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