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Calculate the pH of the solution resulting from the addition of 20.0 mL of 0.100 M NaOH to 30.0 mL of 0.100 M HNO3.

(a) 1.35
(b) 1.70
(c) 1.95
(d) 2.52

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  1. Ph determination

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  2. HNO3 + NaOH ==> NaNO + H2O

    millimols HNO3 = 30 x 0.1 = 3.00
    millimols NaOH = 20 x 0.1 = 2.00
    millimols HNO3 in excess = 1.00

    M HNO3 = mmols/mL = 1.00/50.0 = ?
    Then pH = -log(HNO3) = ?

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  3. 1.70

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