Aspirin is a weak acid with a Ka of 3.0X10-5. Calculate the pH of a solution by

Question

Aspirin is a weak acid with a Ka of 3.0X10-5. Calculate the pH of a solution by dissolving 0.65g of aspirin in water and diluting it to 50.0ml. You may use the symbol RCOOH to represent aspirin when writing chemical reaction. Molecular weight of aspirin = 180.0g/mol Does the pH value that you calculate for the aspirin solution make sense based on the Ka value? Calculate the pH of a solution prepared by mixing 20.00ml of the aspirin solution in the question above with 5.00ml of 0.200M NaOH. Calculate the pH of a solution prepared by mixing 20.00ml of the aspirin solution in the question above with 10.00ml of 0.200M NaOH.

Explanation / Answer

a)

Ka = 3.0 x 10^-5

pKa = 4.52

moles of aspirin = 0.65 / 180 = 3.61 x 10^-3 mol

concentration = 3.61 x 10^-3 / 0.050 = 0.0722 M

RCOOH   ---------------> RCOO-   + H+

0.0722                               0              0

0.0722 - x                           x               x

Ka = x^2 / 0.0722 - x

3 x 10^-5 = x^2 / 0.0722 - x

x = 1.47 x 10^-3

[H+] = 1.47 x 10^-3 M

pH = -log (1.47 x 10^-3 )

pH = 2.83

b)

mmoles of aspirin = 20 x 0.0722 = 1.444

mmoles of NaOH = 5 x 0.2 = 1

RCOOH   +   NaOH   ------------> RCOO-   + H2O

1.444           1.00                           0               0

0.444              0                            1.00

pH = pKa + log [salt / acid]

    = 4.52 + log [1 /0.44]

pH = 4.87

c)

mmoles of NaOH = 10 x 0.2 = 2

RCOOH   +   NaOH   ------------> RCOO-   + H2O

1.444           2.00                           0               0

      0             0.556                       1.44

here strong base remains.

[OH-] = 0.556 / 20+ 10 = 0.0185 M

pOH = -log 0.0185 = 1.73

pH = 12.27

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