Calculate the final pH of a solution made by the addition of 10 mL of a 0.5 M NaOH solution to 500 mL of a 0.4 M HA originally at pH

This question is not complete, the complete question is;

Calculate the final pH of a solution made by the addition of 10 mL of a 0.5 M NaOH solution to 500 mL of a 0.4 M HA originally at pH  = 5.0 ( pKa = 5.0)

Neglect the volume change.

Options:

a) 6.10

b) 5.09

c) 7.00

d) 5.02

the final pH of a solution is 5.02

Option d) 5.02 is the Correct Answer

Explanation:

Given the data in the question,

Initially pKa = pH; so ratio is 1:1

thus, 0.4 M acid and base

Now, moles of NaOH = molarity × volume = 0.5 × 10 = 5 mmol = 5 × 10⁻³ mol.

Going into 500 mL ( 0.5 L ) of solution

new molarity will be;

⇒ moles / volume = 5 × 10⁻³ / 0.5 = 0.01 M

ACID reacting with BASE

original concentration of acid = 0.4 - 0.01 = 0.39 M

original concentration of base = 0.4 + 0.01 = 0.41 M

so

pH = 5 + log( base/acid)

= 5 + log ( 0.41/0.39)

= 5 + log ( 1.0512)

= 5 + 0.021

pH = 5.02

Therefore the final pH of a solution is 5.02

Option d) 5.02 is the Correct Answer

wher is the rest of the question