A solution of potassium chloride has a pH of 7.81.What is the [OH-] in mol/L ?

Can you please show steps I would appreciate it very much.

Thank you

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  1. pH = -log(H^+)
    7.81 = -log*H^+)
    log(H^+) = -7.81
    Take the antilog of both sides.
    (H^+) = 1.5488 x 10^-8 (rounded to 1.55*10^-8)
    (H^+)(OH^-) = 1 x 10^-14
    solve for (OH^-) = 6.46 x 10^-7
    But there is an easier way to do it.

    pH = 7.81
    pH + pOH = 14; therefore,
    pOH = 14 - 7.81 = 6.19
    6.19 = -log(OH^-)
    log(OH^-) = -6.19
    (OH^-) = 6.46 x 10^-7

    Check my work.

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