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
