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23 February, 12:39

Be sure to answer all parts. Calculate the pH during the titration of 30.00 mL of 0.1000 M KOH with 0.1000 M HBr solution after the following additions of acid: (a) 9.00 mL pH = 12.73 (b) 29.80 mL pH = 1 (c) 38.00 mL pH = 1.9

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  1. 23 February, 12:53
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    (a) pH = 12.73

    (b) pH = 10.52

    (c) pH = 1.93

    Explanation:

    The net balanced reaction equation is:

    KOH + HBr ⇒ H₂O + KBr

    The amount of KOH present is:

    n = CV = (0.1000 molL⁻¹) (30.00 mL) = 3.000 mmol

    (a) The amount of HBr added in 9.00 mL of 0.1000 M HBr is:

    (0.1000 molL⁻¹) (9.00 mL) = 0.900 mmol

    This amount of HBr will neutralize an equivalent amount of KOH (0.900 mmol), leaving the following amount of KOH:

    (3.000 mmol) - (0.900 mmol) = 2.100 mmol KOH

    After the addition of HBr, the volume of the KOH solution is 39.00 mL. The concentration of KOH is calculated as follows:

    C = n/V = (2.100 mmol) / (39.00 mL) = 0.0538461 M KOH

    The pOH and pH of the solution can then be calculated:

    pOH = - log[OH⁻] = - log (0.0538461) = 1.2688

    pH = 14 - pOH = 14 - 1.2688 = 12.73

    (b) The amount of HBr added in 29.80 mL of 0.1000 M HBr is:

    (0.1000 molL⁻¹) (29.80 mL) = 2.980 mmol

    This amount of HBr will neutralize an equivalent amount of KOH, leaving the following amount of KOH:

    (3.000 mmol) - (2.980 mmol) = 0.0200 mmol KOH

    After the addition of HBr, the volume of the KOH solution is 59.80 mL. The concentration of KOH is calculated as follows:

    C = n/V = (0.0200 mmol) / (59.80 mL) = 0.0003344 M KOH

    The pOH and pH of the solution can then be calculated:

    pOH = - log[OH⁻] = - log (0.0003344) = 3.476

    pH = 14 - pOH = 14 - 3.476 = 10.52

    (c) The amount of HBr added in 38.00 mL of 0.1000 M HBr is:

    (0.1000 molL⁻¹) (38.00 mL) = 3.800 mmol

    This amount of HBr will neutralize all of the KOH present. The amount of HBr in excess is:

    (3.800 mmol) - (3.000 mmol) = 0.800 mmol HBr

    After the addition of HBr, the volume of the analyte solution is 68.00 mL. The concentration of HBr is calculated as follows:

    C = n/V = (0.800 mmol) / (68.00 mL) = 0.01176 M HBr

    The pH of the solution can then be calculated:

    pH = - log[H⁺] = - log (0.01176) = 1.93
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