Question

Please answer the following questions regarding the titration curve shown below. 

(a) What is true about the analyte and the titrant for this titration? It is a weak acid analyte with a strong base titrant. 

(b) If the titrant has a molarity of 0.3000 M and there are 25.00 mL of analyte present, what is the molarity of the analyte? 

(c) What is the pK_a of the analyte in this titration to the nearest 0.5?

Answer 1

(b) molarity of analyte = 0.228 M

(c) pK_a of analyte = 4.5

Explanation

(b) At equivalence point, volume of titrant added = 19 mL

moles titrant added = (concentration of titrant) * (volume of titrant added)

moles titrant added = (0.3000 M) * (19 mL)

moles titrant added = 5.7 mmol

moles acid = moles base added

moles acid = 5.7 mmol

concentration analyte = (moles acid) / (volume of analyte)

concentration analyte = (5.7 mmol) / (25.00 mL)

concentration analyte = 0.228 M

(c) volume of titrant added at equivalence point = 19 mL

volume of titrant added at half-equivalence point = ( volume of titrant added at equivalence point) / 2

volume of titrant added at half-equivalence point = (19 mL) / 2

volume of titrant added at half-equivalence point = 9.5 mL

From the graph, pH at 9.5 mL is 4.5

pK_a is pH at half-equivalence point

pK_a = 4.5