Titrations Part B A 50.0-mL volume of 0.15 M HBr is titrated with 0.25 M KOH. Ca

Question

Titrations

Part B

A 50.0-mL volume of 0.15 M HBr is titrated with 0.25 M KOH. Calculate the pH after the addition of 18.0 mL of KOH.

Express your answer numerically.

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Part C

A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×105) is titrated with 0.500 M HNO3. Calculate the pH after the addition of 27.0 mL of HNO3.

Express your answer numerically.

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Part D

A 52.0-mL volume of 0.35 M CH3COOH (Ka=1.8×105) is titrated with 0.40 M NaOH. Calculate the pH after the addition of 27.0 mL of NaOH.

Express your answer numerically.

Explanation / Answer

Part B

millimoles of HBr = 50x 0.15 = 7.5

millimoles of KOH = 18 x 0.25 = 4.5

here millimoles of HBr > millimoles of KOH

[H+] = 7.5 - 4.5 / 50 + 18 = 0.044 M

pH = -log (0.044) = 1.355

pH = 1.36

Part C

mmoles of NH3 = 75 x 0.200 = 15

mmoles of HNO3 = 27 x 0.5 = 13.5

NH3   +    HNO3    ---------------> NH4+NO3-

15          13.5                                  0            

1.5             0                                 13.5

pOH = pKb + log [salt / base]

       = 4.74 + log [13.5 / 1.5]

pOH = 5.69

pH = 8.31

part D)

CH3COOH +   NaOH   ------------> CH3COONa

18.2                10.8                               0

7.4                    0                               10.8

pH = 4.74 + log [10.8 / 7.4]

pH = 4.90

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