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Calculate [OH-] and pH for the following strong base solution;

a solution formed by mixing 10.0 mL of 0.011 M Ba(OH)2 with 28.0 mL of 7.4 10-3 M NaOH

working and answer would be appreciated


    mol Ba(OH)2 = M x L = ??.
    (OH^-) from Ba(OH)2 is twice that.
    Add to the following.

    mol NaOH = M x L = ?
    (OH^-) = total mole OH^-/total L

    Then pOH = -log(OH^-)
    Substitute pOH in the following and solve for pH.
    pH + pOH = pKw = 14

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