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

possibly, yes.

explanation:

the amount of sugar you take in is dangerously high, risk of heart disease, high risk of diabetes, and etc.

the positions which points up in upward carbon and down in downward carbon are axial

and vice versa holds true for equatorial positions

so the correct answer with labelled diagram will be