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3 April, 12:08

What is the pH of a solution prepared by dissolving 0.140 g of potassium hydroxide in sufficient pure water to prepare 250.0 ml of solution

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  1. 3 April, 12:14
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    pH = 12

    Explanation:

    Potassium hydroxide (KOH) is a strong base, so it dissociates completely in water by giving OH⁻ anions as follows:

    KOH⇒ K⁺ + OH⁻

    Since dissociation is complete, it is assumed that the concentration of OH⁻ is equal to the initial concentration of KOH:

    [OH⁻] = [KOH]

    In order to find the initial concentration of KOH, we have to divide the mass (0.140 g) into the molecular weight of KOH (Mw):

    Mw (KOH) = K + O + H = 39 g/mol + 16 g/mol + 1 g/mol = 56 g/mol

    moles KOH: mass/Mw = 0.140 g / (56 g/mol) = 2.5 x 10⁻³ moles

    The molality of the solution is the number of moles of KOH per liter of solution:

    V = 250.0 ml x 1 L/1000 ml = 0.250 L

    M = (2.5 x 10⁻³moles) / (0.250 L) = 0.01 M

    Now, we calculate pOH:

    pOH = - log [OH⁻] = - log [KOH] = - log (0.01) = 2

    Finally, we calculate pH from pOH:

    pH + pOH = 14

    ⇒pH = 14 - pOH = 14 - 2 = 12
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