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
Answer:
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
