Question

3. In the lab you are going to titrate a weak acid solution (25.00 mL of 0.100 M HCHO2, formic acid) with a strong base (0.100 M NaOH). Carry out the following calculations to determine the pH for four key points throughout the titration.

Part a. Calculate the volume of 0.100 M NaOH required to reach the equivalence point for the titration.   Ans. 25.00 mL

Part b. Calculate the initial pH of 0.100 M HCHO2 (before adding any NaOH).   Ans. pH = 2.37

Part c. Calculate the pH after the addition of 5.00 mL of the NaOH solution.   Ans. pH = 3.16

Part d. Calculate the pH after the addition of 12.50 mL of the NaOH solution (half-equivalence point).   Ans. pH = 3.74

Part e. Calculate the pH after the addition of 25.00 mL of the NaOH solution (equivalence point).   Ans. pH = 8.22

Part f. Calculate the pH after the addition of 40.00 mL of the NaOH solution.   Ans. pH = 12.3632

Answer 1

Answer:-

The answer is given in the image,

[HCHO_2] = 0.100 M [Na OH] = 0.100 M (a) Moles of Na OH = Moles of HCHO_2 so, volume of Na OH = 0.100 times 25.00/0.100 = 25.00 m_2 (b) HCHO_2 + H_2O CHO^-_2 + H_3 O^+ ka = [CHO_2^-][H_3 O^+]/[H CHO_2] 1.8 times 10^-4 = x.x/0.100 - x 1.8 times 10^-5 - 1.8 times 10^-4 x = x^2 x^2 + 1.8 times 10^-4 - 1.8 times 10^-5 = 0 x = -1.8 times 10^-4 + squareroot (1.8 times 10^-4)^2 + 4 times 1.8 times 10^-5/2 \r\n x = -1.8 times 10^-4 + 8.49 times 10^-3 x = 4.15 times 10^-3 [H_3 O^-] = 4.15 times 10^-3 P^H = -log (4.15 times 10^-3) P^H = 2.38 C) After addition y 5.0 ml Moles of Na OH = Moles CHO_2^- = 0.100 times 5.00/1000 = 0.500 times 10^-3 Moles of HCHO_2 = 0.100 times 25/1000 = 2.5 times 10^-3 mol