Question

calculate the pH of a 0.01 N acid, completely dissociated.

calculate the pH of a 0.002 N base, complete ly dissociated.

calculate the pH of a 0.05 M weak acid that is 2% disassociated.

the pH of a solution is 2.3 calculate the [H+]

when 20 mL of 0.5 M solution of a weak acid are mixed with 5.0 mL of 0.5 M NaOH, the pH is found to be 4.75. what is the pKa of the acid?

Answer 1

calculate the pH of a 0.01 N acid, completely dissociated.

For a monobasic acid,

Concentration of acid = 0.01 N

As it completely dissociated, [H⁺] = 0.01 mol dm^-3

pH = - log[H⁺] = - log[0.01] = 2

calculate the pH of a 0.002 N base, complete ly dissociated.

For a monoacidic base,

Concentration of base = 0.002 N

As it completely dissociated, [OH^-] = 0.002 mol dm^-3

pOH = - log[OH^-] = - log[0.002] = 0.301 * 3= 0.903

pH = 14 - 0.903 = 13.097

calculate the pH of a 0.05 M weak acid that is 2% disassociated.

Strength of acid is 0.05 M, and only 2% dissociates

Thus, [H⁺] = 0.05 * 2% M = 0.001 M

pH = - log[H⁺] = - log[0.001] = 3

the pH of a solution is 2.3 calculate the [H+]

pH = -log [H⁺] = 2.3

=> [H] = 10^-2.3 = 0.005 M

when 20 mL of 0.5 M solution of a weak acid are mixed with 5.0 mL of 0.5 M NaOH, the pH is found to be 4.75. what is the pKa of the acid?

Numer of millimole of base in the solution = 5 * 0.5 = 2.5

Final volume is 5 + 20 = 25 mL

So, concentration of base, [OH^-] = (2.5*1000)/(25*1000) mol dm^-1 = 0.1 M

pOH = -log[OH^-] = -log[0.1] = 1

So, pH for NaOH is 13.

The final pH is 4.75. Thus acid exerts a pH of, 13 - 4.75 = 8.25

Number of millimole of acid in the solution = 20 * 0.5 = 10

So, concentration of acid present, = (10*1000)/(25*1000) mol dm^-1 = 0.4 M

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

=> 8.25 = pKa + (-8.25) - (-0.602) = pKa - 7.648

=> pKa = 15.89

Its a very weak acid or strong base.