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

i'm pretty sure its b i might be wrong though!

the answer to your question is c