A 20.0-mL sample of 0.150 M KOH is titrated with 0.125 M HClO4 solution. Calcula
ID: 1050809 • Letter: A
Question
A 20.0-mL sample of 0.150 MKOH is titrated with 0.125 MHClO4 solution. Calculate the pHafter the following volumes of acid have been added.
Part A
20.0 mL
Express your answer using two decimal places.
______
Part B
23.5 mL
Express your answer using two decimal places.
Part C
24.0 mL
Express your answer using two decimal places.
Part D
26.5 mL
Express your answer using two decimal places.
Part E
32.0 mL
Express your answer using two decimal places.
A 20.0-mL sample of 0.150 MKOH is titrated with 0.125 MHClO4 solution. Calculate the pHafter the following volumes of acid have been added.
Part A
20.0 mL
Express your answer using two decimal places.
pH =______
Part B
23.5 mL
Express your answer using two decimal places.
pH = ______Part C
24.0 mL
Express your answer using two decimal places.
pH = ______Part D
26.5 mL
Express your answer using two decimal places.
pH = ________Part E
32.0 mL
Express your answer using two decimal places.
pH = _______Explanation / Answer
millimoles of KOH = 20 x 0.15 = 3.0
A) millimoles of HClO4 added = 20 x 0.125 = 2.5
3.0 - 2.5 = 0.5 millimoles NaOH left
[NaOH] = 0.5 / 40 = 0.0125 M
pOH = - log [OH-]
pOH = - log [0.0125]
pOH = 1.903
pH = 14 - 1.903
pH = 12.09
B) millimoles of HClO4 = 23.5 x 0.125 = 2.937
3 - 2.937 = 0.063 millimoles NaOH left
[NaOH] = 0.063 / 43.5 = 0.00145 M
pOH = -log [OH-]
pOH = - log [0.00145]
pOH = 2.84
pH = 14 - 2.84
pH = 11.16
C ) millimoles of HClO4 = 24 x 0.125 = 3.0
at equivalence point
pH = 7.0
D) millimoles of HClO4 = 26.5 x 0.125 = 3.312
3 - 3.312 = 0.312 millimoles HClO4 left
[HClO4] = 0.314 / 46.5 = 0.0067 M
pH = - log [H+]
pH = - log [0.0067]
pH = 2.17
E) millimoles of HClO4 = 32 x 0.125 = 4.0
4 - 3 = 1.0 millimoles HClO4 left
[HClO4] = 1 / 52 = 0.019 M
pH = - log [H+]
pH = - log [0.019]
pH = 1.72
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