Question

please include how you solved the answer

N Le 2.30u is. You have 500.0 mL of a buffer solution containing 0.30 M acetic acid (CH2COOH) and 0.25 M sodium acetate (CH,COONa). What will the pH of this solution be after the addition of 20.0 mL of 1.00 M NaOH solution? [K,= 1.8 x 10-51 DKA - 109( 1.8x10-5) -4.74 a. 4.88 4.74 +log .30 = 4.8 b. 4.74 c. 4.67 d. 4.79 4.8 X e. 4.54 25

Answer 1

mol of NaOH added = 1.0M *20.0 mL = 20.0 mmol

CH3COOH will react with OH- to form CH3COO-

Before Reaction:

mol of CH3COO- = 0.25 M *500.0 mL

mol of CH3COO- = 125 mmol

mol of CH3COOH = 0.3 M *500.0 mL

mol of CH3COOH = 150 mmol

after reaction,

mol of CH3COO- = mol present initially + mol added

mol of CH3COO- = (125 + 20.0) mmol

mol of CH3COO- = 145 mmol

mol of CH3COOH = mol present initially - mol added

mol of CH3COOH = (150 - 20.0) mmol

mol of CH3COOH = 130 mmol

Ka = 1.8*10^-5

pKa = - log (Ka)

= - log(1.8*10^-5)

= 4.745

since volume is both in numerator and denominator, we can use mol instead of concentration

use:

pH = pKa + log {[conjugate base]/[acid]}

= 4.745+ log {1.45*10^2/1.3*10^2}

= 4.792

Answer: d