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13 Dec 2019

A 20.00-mL sample of a 0.200 M weak acid HA is titrated with 0.400 M NaOH as the titrant. At the equivalence point, the pH of the sample is 8.71. What is Ka of the weak acid?

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Bunny Greenfelder
Bunny GreenfelderLv2
17 Dec 2019

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Related questions

A 20. mL sample of a 0.10 M weak acid (KA= 1.3×10−4 ) is titrated with 0.20 M NaOH. If 10.0 mL of NaOH are needed to reach the equivalence point, what is the pH after addition of 5.0 mL of NaOH (halfway to the equivalence point)?

a. 10.0

b. 8.71

c. 5.35

d. 4.57

e. 3.89

Aniline hydrochloride, [C6H5NH3]Cl, is a weak acid. Its conjugate base is the weak base aniline, C6H5NH2. Ka= 2.40 X 10-5.

C6H5NH3+(aq) + OH-(aq) → C6H5NH2(aq) +H2O(l)

Assume 50.0 mL of 0.100 M aniline hydrochloride is titrated with 0.200 M NaOH. What is the pH at the equivalence point?

A lab experiment was conducted utilizing software to general a titration curve. 25 mL of an unknown analyte was titrated with .1015 M of NaOH solution.

The pH at the equivalence point for solution 1 = 7.83.

12.61mL of NaOH titrant added.

The pH at the equivalence point for solution 2 = 8.53. Total volume of NaOH titrants for solution 1

12.47mL of NaOH titrant added.

What is Ka for the weak acid?

Please help with any calculations here.

Part 2 consists of unknown acid solutions with added base in which the pH was measured. Please help calculate the [A-]/[HA] of buffers.

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