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28 July, 15:05

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

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  1. 28 July, 15:09
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    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
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