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12 April, 07:03

A chemist boils off the water from 627 mL of a solution with a pH of 10.2. The only dissolved compound in the solution is NaOH. How much solid remains

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  1. 12 April, 07:19
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    4.03 x 10∧-3

    Explanation:

    The pH of the solution is 10.2, and we know:

    pH + pOH = 14

    pOH = 14 - pH

    pOH = 14 - 10.2 = 3.8

    pOH = - log[OH-]

    3.8 = - log[OH-]

    -3.8 = log[OH-]

    [OH-] = 10∧-3.8 = 1.58 * 10∧-4

    since NaOH (s) → Na∧ + (aq) + OH∧ - (aq), [NaOH] = 1.58*10-4.

    627mL * 1L/1000mL * 1.58*10-4molesNaOH/1L * 40g NaOH/1mole NaOH

    =4.03 * 10∧-3g NaOH
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