Need Help. Thanku A 20.0-mL sample of 0.150 M KOH is titrated with 0.125 M HClO4

Question

Need Help. Thanku

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.

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Part B

22.0mL

Express your answer using two decimal places.

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Part C

24.0 mL

Express your answer using two decimal places.

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Part D

26.0mL

Express your answer using two decimal places.

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Part E

32.0mL

Express your answer using two decimal places.

Explanation / Answer

I think the other answer misread part of this question.

In your 20mL sample of 0.150 M KOH you will have 0.150 * 20/1000 = 3.000 * 10^-3 mole in 20 mL of solution.

After each titration stage you have added the following amount of HClO4:
1. (20/1000) * 0.125 = 2.50 * 10^-3 mole
2. (22.5/1000) * 0.125 = 2.81 * 10^-3 mole
3. (24/1000) * 0.125 = 3.00 * 10^-3 mole
4. (26.5/1000) * 0.125 = 3.31 * 10^-3 mole
5. (32/1000) * 0.125 = 4.00 * 10^-3 mole.

Your reaction is
KOH + HClO4 <==> KClO4 + H2O

Solution 1. [OH^-] is (3.000 * 10^-3 - 2.50 * 10^-3) *1000/(20+20) = 0.0125 M
Solution 2. [OH^-] is (3.000 * 10^-3 - 2.81 * 10^-3) * 1000/(20+22.5) = 0.00447 M
Solution 3. is neutral, so the PH is 7.
Solution 4. [H^+] is (3.31 * 10^-3 - 3.000 * 10^-3) * 1000/(20+26.5) = 0.00667 M
Solution 5 [H^+] is (4.00 * 10^-3 - 3.000 * 10^-3) * 1000/(20+32) = 0.0192 M

We know pH = -log[H^+], pOH = -log[OH^-] and that PH = 14.0 - pOH
Solution 1. pOH = 1.903, therefore PH = 12.09
Solution 2. pOH = 2.350, therefore PH = 11.65
Solution 3 pH = 7.000
Solution 4 pH = 2.175
Solution 5. pH = 1.717.

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