Concentration of acetic acid: 1.041M and concentration of sodium acetate is 1.07

ID: 956268 • Letter: C

Question

Concentration of acetic acid: 1.041M and concentration of sodium acetate is 1.071M.

The buffer is created by mixing 100mL of acetic acid with 100mL of sodium acetate soluiton.

a) What is the pH of 75mL of the buffer solution after the addition of 1mL of 3M NaOH?

b) What is the pH of 75mL of the buffer solution after the addition of 5mL of 3M NaOH?

c) What is the pH of 75mL of the buffer solution after the addition of 15mL of 3M NaOH?

d) Calculate the volume of 3M HCl needed to change the pH of 75mL of the buffer solution by one pH unit (buffer capacity).

PLEASE SHOW ALL WORK. I DID SOME MISTAKES SOMEWHERE AND WOULD LIKE TO SEE FULL SOLUTIONS TO COMPARE.

Thank you very much.

a ) Calculate concentration of CH3COONa & CH3COOH separately in 200ml of buffer solution using

M1V1 = M2V2 , & then substitute the values in Handerson's equation to find out pH of buffer

solution. The pKa value of acetic acid is known to be equal to 4.7447

thus , [CH3 COOH ] = { (1.041 x 1000) / 200 } = 5.205 M

similarly [CH3COONa] = { ( 1.071 x 1000 ) / 200 } = 5.355 M

Handerson - Hasselbalch's equation ..........pH = pKa + log ( [ salt ] / [ acid ] )

.......................substituting the values ,...............= 4.7447 + log { 5.355 / 5.205 }

........................................................................... = 4.7570

................................................................or, ......~= 4.76

Now, 1ml of 3M NaOH contains NaOH = ( 3.0 / 1000 ) =0.003 moles of NaOH

So, 0.003 moles of NaOH yields 0.003 moles of OH- ions . These combine with 0.003 moles of H+ ion

to form H2O and the equilibrium

..........................................CH3COOH <------------> H+ + CH3COO-

is shifted to the right (LeChatelier principle ). Thereby 0.003 moles of CH3COOH dissociate to form

0.003 mole of H+ needed to neutralize OH- ions and also 0.003 moles of CH3COO- ions. Hence ,

the concentration of CH3COOH is decreased by 0.003 moles / L and CH3COO- is increased by 0.003 moles / L

.........................CH3COOH <---------------> H+ + CH3COO-

.........................(1.041- 0.003 ).......................................(1.071 + 0.003 )

So, [ CH3COOH ] = 1.038 M...................& [ CH3COO- ] = 1.074

The equilibrium expression may now be written as,

.....................Ka* = [ H+ ] [ CH3COO- ] / [CH3COOH]

................. 1.8 x 10-^-5= ( [ H+ ] x 1.074 ) / 1.038

therefore [ H+ ] = 1.7397 x 10^-5

& pH = - log [ H^+ ] = - log ( 1.7397 x 10^-5 )

....................................... = 4.7595

Note,.................... Ka* for CH3COOH = 1.8 x 10^-5

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Glad to help. Please post the other questions separately for detailed answers.

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