Calculate the pH of a buffer solution after the addition of 15.00 mL of 0.100 M

Question

Calculate the pH of a buffer solution after the addition of 15.00 mL of 0.100 M NaOH to 50.0 mL of the buffer that originally contains 0.100 M propanoic acid and 0.0750 M sodium propanoate. The K_a for propanoic acid is 1.3 Times 10^-5. Will a precipitate form if 100.0 mL of 2.5 Times 10^-3M Cd(NO_3)_2 and 75.0 mL of 0.0500 M NaOH are mixed at 25 degree C? Calculate the concentration of cadmium ion in solution at equilibrium after the two solutions are mixed. (K_sp is 7.2 Times 10^-15 for Cd(OH)_2 at 25 degree C)

Explanation / Answer

M = 0.1 NaOh

V = 15 ml

m = 0.1 prop acid

M = 0.075 sodium prop

Ka = 1.3*10^-5

pKa = -log(ka) = -log( 1.3*10^-5) = 4.886

The buffer

pH = pKa + log(prop - / prop acid)

base added = M*V = 15*.1 = 1.5 mmol of NaOH

initial acid = M*V = 0.1*50 = 5 mmol of prop acid

initial prop- = M*V = 0.075*50 = 3.75 mmol of prop

then, after reaction takes place:

acid final = 5 - 1.5 = 3.5

prop - final = 3.75+1.5 = 5.2

Then

pH = pKa + log(prop - / prop acid) = 4.886 + log(5.2/3.5) = 5.057

pH = 5.06

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