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What should be the pH of a 0.1M NaOH solution?

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  • 1 decade ago
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    pH = -log [H+]

    Where [H+] is the concentration of the H+ ions present in the solution.

    pOH = -log [OH-]

    Where [OH-] is the concentration of the OH- ions present in the solution.

    pH + pOH = 14

    So, pH = 14 – pOH = 14 – (-log [OH-])

    We are given the concentration of an NaOH solution. NaOH is a strong electrolyte and will disassociate completely into Na+ and OH- ions in solution. These OH- ions from the NaOH will contribute to raise the pH (and lower the pOH) of the solution.

    For every 1 mole of NaOH dissolved in solution, 1 mole of OH- ions are releases (one to one ratio). For if you have a .1 Molar NaOH solution, you have a .1 Molar OH- solution.

    Since the concentration of the NaOH solution given in the problem is MUCH greater than the natural OH- concentration in the water due to water’s tendency to disassociate with itself (H2O --> H+ + OH-), we can neglect the water’s contribution to the [OH-] and focus solely on the NaOH’s contribution to the [OH-].

    The pH of pure water is 7, meaning that it disassociates into H+ and OH and the equilibrium lies where [H+] = [OH-] = 10^-7 Molar = .0000001 Molar.

    .1 Molar >> .0000001 Molar, so we are justified in making this estimation, the [OH-] due to the NaOH is 1 million times greater than due to the water.

    [NaOH] = [OH-] = .1 Molar

    pOH = -log [OH-] = -log [NaOH] = -log (.1) = 1

    pOH = 1

    pH = 14 – pOH = 14 – 1 = 13

    pH = 13

    So the pH of a .1 Molar solution of Sodium Hydroxide is 13.

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  • 1 decade ago

    OH is a strong base

    [OH] = .1

    pOH = -log[OH]

    pOH = -log[.1] = 1

    pH = 14- pOH

    pH = 13

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  • 1 decade ago

    13.0

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