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A biochemist dissolved 5.00 g citric acid in 250 mL of water. The pKa of citric...

A biochemist dissolved 5.00 g citric acid in 250 mL of water. The pKa of citric acid is 3.13 and the molecular weight is 192 g/mol. Determine the pH of the solution and the ratio of conjugate base to weak acid at equilibrium.

Expert Solution

moles of acid = mass / molar mass

= 5.00 / 192

= 0.0260

volume = 0.250 L

molarity = moles / volume

= 0.0260 / 0.250

= 0.104 M

acid = HA

HA <----------------------------> H+ + A-

0.104 0 0

0.104-x x x

Ka = x^2 / 0.104 -x

10^-3.13 = x^2 / 0.104 -x

7.41 x 10^-4= x^2 / 0.104 -x

x = 8.42 x 10^-3

[H+] = 8.42 x 10^-3 M

pH = -log [H+]

pH = -log (8.42 x 10^-3)

pH = 2.07

conjugate base = 8.42 x 10^-3

acid = 0.104 - (8.42 x 10^-3) = 0.0956 M

ratio of conjugate base / weak acid = 8.42 x 10^-3 /0.0956

= 0.0881

queen_honey_blossom answered 2 years ago

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