CHM1 17Example1 Collection

Example 1

Worked Problem Here

Examples 1, 2, and 3 are all in this one Livescribe recording

Example 1 Solution

Because 1.0 X 10^-14 = [H₃O⁺][OH^-],
To find the concentration of H₃O⁺, solve for the [H₃O⁺].

(1.0 X 10^-14)/[OH^-] = [H₃O⁺]
(1.0 X 10^-14)/(2.5 X 10^-4) = [H₃O⁺] = 4.0 X 10^-11 M

Example 2 Solution

  1.   The equation for pH is -log [H+]

    [H⁺]= 2.0 X 10^-3 M
    pH = -log [2.0 X 10^-3] = 2.70 

2.  The equation for pOH is -log [OH^-]

    [OH^-]= 5.0 X 10^-5 M
    pOH = -log [5.0 X 10^-5] = 4.30
    pK_w = pH + pOH and pH = pK_w - pOH
    Then pH = 14 - 4.30 = 9.70.

Example 3 Solution

pH = -log [H⁺]
7.84 = -log [H⁺]
[H+] = 1.45 x 10^-8 M

(Hint: place -7.84 in your calculator and take the antilog (often inverse log or 10^x) = 1.45 x 10^-8M
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