What is the ph of a solution after 35ml of 0.04m naoh was added to 250ml of deionized water?

Answer is 11.69.

Explanation;

We can use two formulas to find out the pH of the NaOH solution,

pOH = - log[OH⁻ (aq) ] (1)

pH + pOH = 14 (2)

First find out the pOH.

To find out the pOH, we should calculate the final concentration of OH⁻ ions.

NaOH is a strong base. Hence, it fully dissociates into ions as Na⁺ and OH⁻.

NaOH (aq) → Na⁺ (aq) + OH⁻ (aq)

Molarity = number of moles of the solute (mol) / volume of the solution (L)

Initial NaOH solution

Molarity of NaOH = 0.04 mol/L

Volume of NaOH = 35 mL = 35 x 10⁻³ L

Hence, moles of NaOH = Molarity x volume

= 0.04 mol/L x 35 x 10⁻³ L

= 1.4 x 10⁻³ mol

The 250 mL of deionised water was added to 35 mL of NaOH.

Then,

new volume of the solution = 35 mL + 250 mL

= 285 mL

The initial amount of NaOH (1.4 x 10⁻³ mol) is in now 285 mL of the solution.

Hence,

Molarity = 1.4 x 10⁻³ mol / 285 x 10⁻³ L

= 4.912 x 10⁻³ mol/L

The stoichiometric ratio between NaOH and OH⁻ is 1 : 1.

Hence,

the molarity of OH⁻ = 4.912 x 10⁻³ mol/L

Hence, pOH = - log (4.912 x 10⁻³ mol/L)

= 2.31

Now, pH can be calculated by using (2) formula.

pH + 2.31 = 14

pH = 14 - 2.31

pH = 11.69