If 20 mL of a 0.1 M NaOH is added to 30 mL of 0.2M CH3COOH (pka=4.74), the pH of the resulting solutiin is
If 20 mL of a 0.1 M NaOH is added to 30 mL of 0.2M CH3COOH (pka=4.74), the pH of the resulting solutiin is
Number of gram equivalents(eq) = moles x acidity of a base (or bascity of an acid.)
Or
= Molarity x Volume in litres x acidity of a base (or bascity of an acid)
So,
Eq(NaOH) = 20/1000 x 0.1 x 1 = 0.002 eq
Eq(CH3COOH) = 30/1000 x0.2 x1 = 0.006 eq
That means acid = base. Hence a salt would be formed in the solution with total volume = 30 + 20 =50 ml
The pH of a strong base weak acid salt solution = 7 + 0.5(pKa +logC) where C is concentration of salt
concentration of salt = eq/volume =1000 x 0.002/50 = 0.4
So
pH = 7 + 0.5(4.74 +log0.4)
pH = 9.171
For such questions always first determine what will exist after the mixture has reacted among itself. Once you know what is present, you can apply the formula.
Number of gram equivalents(eq) = moles x acidity of a base (or bascity of an acid.)
Or
= Molarity x Volume in litres x acidity of a base (or bascity of an acid)
So,
Eq(NaOH) = 20/1000 x 0.1 x 1 = 0.002 eq
Eq(CH3COOH) = 30/1000 x0.2 x1 = 0.006 eq
That means acid = base. Hence a salt would be formed in the solution with total volume = 30 + 20 =50 ml
The pH of a strong base weak acid salt solution = 7 + 0.5(pKa +logC) where C is concentration of salt
concentration of salt = eq/volume =1000 x 0.002/50 = 0.4
So
pH = 7 + 0.5(4.74 +log0.4)
pH = 9.171
For such questions always first determine what will exist after the mixture has reacted among itself. Once you know what is present, you can apply the formula.
