Lab Experiment 11 Acid‐Base III – pH Indicators, Salts, and Buffers 4. Experimentally, acetate buffer HC2H3O2‐C2H3O2...

Lab Experiment 11 Acid‐Base III – pH Indicators, Salts, and Buffers

4. Experimentally, acetate buffer HC2H3O2‐C2H3O2 ‐ can be prepared in different ways.

a. Calculate the approximate pH of an acetate buffersolution prepared by mixing equal volumes of 0.10 M HC2H3O2 and 0.10 M NaC2H3O2 (Ka, HC2H3O2 = 1.8 x 10‐5 ). Show your work.

b. If an acetate buffer solution was going to be prepared by neutralizing HC2H3O2 with 0.10 M NaOH, what volume (in mL) of 0.10 M NaOH would need to be added to 10.0 mL of 0.10 M HC2H3O2 to prepare a solution with pH = 5.5 ? Show your work.

Solutions

Expert Solution

Buffer formula

pH = pKa + log [Salt] / [Acid]

pKa = -log Ka = - log [ 1.8 x 10‐5 ] = 4.744

pH = 4.744 + log [0.1] / [0.1]

pH = 4.744

Volume doesnt effect the pH value since we are using salt to acid ratio

So answer ---> pH = 4.744

pH = pKa + log [ CH3COONa] / [CH3COOH]

5.5 = 4.744 + log [CH3COONa] / [CH3COOH]

[CH3COONa] / [CH3COOH] = 5.7

x/y = 5.7

x = 5.7y

Moles of CH3COOH = Molarity*V in L = 1 milli mole = 10^-3 mole

MOles of NaOH = 0.1*V

CH3COOH + NaOH -------> CH3COONa

1*10^-3 0.1V 0

10^-3 - 0.1V 0 0.1V

0.1V = 5.7 (10^-3 - 0.1V)

0.1V = 5.7*10^-3 - 0.57V

0.67V = 5.7*10^-3

V = 8.5 ml

VOlume of NaOH needed = 8.5 ml

queen_honey_blossom answered 1 year ago

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