Chemistry

A 635 mL solution of HBr is titrated with 0.51 M KOH. If it takes 1569 mL of the base solution to reach the equivalence point, what is the pH when only 141 mL of the base has been added to the solution?

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  1. HBr + KOH ==> KBr + H2O
    mols KOH = M x L = 1.569 x 0.51 approx 0.8.
    mols HBr = 0.8 (look at the coefficients in the balanced equation; 1 mol HBr to 1 mol KOH.)
    M HBr = mols HBr/L HBr = approx 0.8/0.635 = approx 12.6M

    mols HBr initially = 0.8
    mols KOH added with 141 mL = M x L = 0.51 x 0.141 = about 0.07
    mols HBr remaining = -.8-0.7 = about 0.7.
    Then (OH^-) = about 0.7
    pOH = -log (OH^-)
    and pH + pOH = pKw = 14.
    You know pKw and pOH, solve for pH. Post your work if you get stuck. NOTE: I've estimated above. You must recalculate each step.

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