In our lab class, we titrated 0.0750 M NaOH in to 10.00 mL of an unknown weak ac

Question

In our lab class, we titrated 0.0750 M NaOH in to 10.00 mL of an unknown weak acid mixed with 40 mL of deionized water. We measured the pH after every 0.5 mL in order to create a titration curve. I have posted our results above. Please tell me the equivalence point, and the value of Ka.

The unknown can be one of these 6 acids, based on what the Ka value is:

Acetylsalicylic acid- 3.3x10^-4 Ka value

Lactic acid- 1.4x10^-4 Ka value

Benzoic acid- 6.5x10^-5 Ka value

Potassium hydrogen phthalate- 3.9x10^-6 Ka value

Hydrogen chromate ion- 3.0x10^-7 Ka value

Phenol- 1.3x10^-10 Ka value

After determining the identity of the unknown acid based on the chart right above, how can I calculate the original molarity of the unknown acid?

Explanation / Answer

the pH = 11.16 ,

Salt concentraion = millimoles of salt /total voulme = 0.075x15/65 = 0.01730

so at the end of titration, only the salt of the acid would be present.

its a salt of strong base and weak base , so

pH = 7 +1/2(pka)+1/2(log(salt conc)

11.16 = 7 +1/2(pka)+1/2log(0.01730 )

pKa = 8.32 + 1.76 = 10.08

so its phenol which has Phenol- 1.3x10^-10 Ka value

The molarity can be calulated by law equivalents as follows :

Milliequvalents of acid = milliequvalents of base added

0.075x15 = 10xMacid

Macid = .1125 M approx

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