Calculate the pH and percentage ionization of a 0.0500 M acetic acid (CH_3COOH)

Question

Calculate the pH and percentage ionization of a 0.0500 M acetic acid (CH_3COOH) solution? Caffeine (C_8H_10N_4O_2) is a weak base with a Kb of 4.10 times 10^-4. Calculate the pH and % ionization of a solution containing a caffeine concentration of 455 mg/mL. A 0.0115 M solution of a weak acid has a pH of 3.29. Calculate the acid ionization constant (Ka) for the acid. A 0.085 M solution of monoprotic acid has a percent ionization of 0.59%. Determine the acid ionization constant (Ka) for the acid. A 0.150M solution of a weak base morphine has a pH of 10.5. What is the Kb of morphine?

Explanation / Answer

1.

CH3COOH (aq) = CH3COO- (aq)+ H+ (aq)

show the ICE table using molarity

CH3COOH (aq) = CH3COO- (aq)+ H+ (aq)

          I                  0.0500M             0                 0

C                -x                        +x               +x

E                 0.0500M-x                   x                 x

here pka= 4.79

ka = 10^-pKa= 10^-4.79

= 1.6*10^-5

show the equilibrium expression here(k) and calculate the value of the unknown for the ice table

Now Ka = x*x/0.0500-x

1.6*10^-5 = x^2 / 0.0500-x

8.0*10^-7 - 1.6*10^-5 x = x^2

X^2 + 1.6*10^-5 x -8.0*10^-7= 0 solve this equation we will get

x= 0.000886M

OR H+ = 0.000886M

pH = - log [H+]

= - log [0.000886M]

= 3.05

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