A 75.0-ml volume of 0.200 m nh3 (kb=1.8*10-5) is titrated with 0.500 m hno3. calculate the ph after the addition of 19.0 ml of hno3. express your answer numerically.

Question

Jamarcus Fischer

Chemistry

8 September, 20:15

A 75.0-ml volume of 0.200 m nh3 (kb=1.8*10-5) is titrated with 0.500 m hno3. calculate the ph after the addition of 19.0 ml of hno3. express your answer numerically.

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Answers (1)

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Rebecca Maxwell · Answer 1

8.03.

Explanation:

Since, the no. of millimoles of NH₃ is more than that of HNO₃, the medium will be basic.

C of base (NH₃) = [ (MV) NH₃ - (MV) HNO₃] / Vtotal.

C of base (NH₃) = [ (0.20 M) (75.0 mL) - (0.50 M) (19.0 mL) ] / (94.0 mL) = 0.0585 M.

∵ [OH⁻] = √ (Kb. C)

∴ [OH⁻] = √ (1.8 x 10⁻⁵) (0.0585 M) = 1.053 x 10⁻⁶.

∵ pOH = - log[OH⁻] = - log (1.053 x 10⁻⁶) = 5.97.

∴ pH = 14 - pOH = 14 - 5.97 = 8.03.