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Title: A 20.0-mL sample of 0.150 M KOH is titrated with 0.125 M HClO4 solution. Calcula Post by: hc23881 on Mar 25, 2013 A 20.0-mL sample of 0.150 M KOH is titrated with 0.125 M HClO4 solution. Calculate the pH after the following volumes of acid have been added.
(a) 21.0 mL . (b) 23.0 mL 24.0 mL (Express your answer to 2 decimal places.) 28.0 mL 31.0 mL Title: Re: A 20.0-mL sample of 0.150 M KOH is titrated with 0.125 M HClO4 solution. Calcula Post by: bio_man on Mar 25, 2013 Similar question A 30.0 mL sample of 0.150 M KOH is titrated with 0.125 M HClO4 solution. Calculate the pH after the following volumes of acid have been added: A) 30.0 mL B) 34.0 mL C) 36.0 mL D) 37.0 mL E) 43.0 mL Solutions 30.0 mL x 0.150 M KOH = 4.50 millimoles KOH (a strong base) 0.125 M HClO4 (a strong acid) NaClO4 will be neutral (A) 30.0 mL x 0.125 M HClO4 = 3.75 millimoles HClO4 4.50 - 3.75 = 0.75 millimoles KOH remain in 60 mL solution 0.75 mmole/60 mL = 0.0125 M KOH pOH = 1.90; pH = 12.10 (B) 34.0 mL x 0.125 M HClO4 = 4.25 millimoles HClO4 4.50 - 4.25 = 0.25 millimoles KOH remain in 64 mL solution 0.25 mmole/64 mL = 0.00391 M KOH pOH = 2.41; pH = 11.59 (C) 36.0 mL x 0.125 M HClO4 = 4.50 millimoles HClO4 4.50 - 4.50 = 0.00 millimoles remain in 65 mL solution pOH = 7.00; pH = 7.00 (D) 37.0 mL x 0.125 M HClO4 = 4.625 millimoles HClO4 4.625 - 4.50 = 0.125 millimoles HClO4 remain in 66 mL solution 0.125 mmole/66 mL = 0.00189 M HClO4 pH = 2.72 (E) 43.0 mL x 0.125 M HClO4 = 5.375 millimoles HClO4 5.375 - 4.50 = 0.875 millimoles HClO4 remain in 73 mL solution 0.875 mmole/73 mL = 0.0120 M HClO4 pH = 1.92 Title: Re: A 20.0-mL sample of 0.150 M KOH is titrated with 0.125 M HClO4 solution. Calcula Post by: hc23881 on Mar 25, 2013 thank you i have gotten the ones before the equivolence point but can not get d or e?!
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