Using the standardized NaOH solution from question 1, the student titrates exact

Question

Using the standardized NaOH solution from question 1, the student titrates exactly 10.0 an unknown monoprotic weak acid as was done in this experiment and obtains the follo Volume at equivalence point: Volume at one half equivalence point: 6.3 mL pH at the one half equivalence point:4.72 a. What is the pKa of the weak acid used in this experiment? 2. 12.6 mL What is the Ka of the weak acid used in this experiment? Show your calculations her full credit. b. c. Based on the pKa value, what could the weak acid be? CHEM 1412- Acid Dissociation Constant, Ke (10/17 How many moles of NaOH were needed to reach the equivalence point in this titratic Show your calculations here for full credit. d. How many moles of the weak acid are present in the 10.0-mL aliquot used? e. f. What is the molarity (mol/L) of the weak acid? Show your calculations here for full

Explanation / Answer

a.titration between strong base and weak acid will prepared a buffer solution.

so, pH = pKa + log {[salt]/[acid]}

at half equivalence point [salt] = [acid]

therefore, pH at half equivalence point will be = pKa

so, pKa = 4.72.

b)pKa = -log Ka

or, Ka = 10-pKa = 10 -4.72  = 1.90*10-5

c) acetic acidity

d) moles = molar* solution in Liter

note: there is no information regarding the strength of NaOH

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