Question

To determine the Ca^2+ concentration in water sample, a standard EDTA solution of 0.01988 M was used to titrate 25 ml. of the sample solution with the presence of an ammonium buffer (pH 10). If 15.80 ml. of the standard EDTA was used to reach the end point, calculate the molar concentration and the ppm concentration of the unknown Ca^2+ (atomic weight of Ca = 40.08g/mole) solution The amount of 0.2915g of benzoic acid was dissolved in 100 ml. of water and titrated by a standard solution of 0.1106 M NaOH (FW=122.12g/mole.K_s = 6.3 Times 10^-5). Calculate the following:) Solution pH before titration starts solution pH half may to equivalence point? solution pH at equivalence point? solution pH at the point where a total at 23.50 mL of NaoH has been added

Answer 1

3 )

Ca+2 + EDTA ----------------> CaEDTA

millimoles of EDTA = millimoles of Ca+2

0.01988 x 15.80 = C x 25

C = 0.01256 M

molar concentration of Ca+2 = 0.01256 M

molarity = mass (g) / molar mass ) x 1/ V (L)

0.01256 mole/ litre = 0.01256 x 40.08 x g / L

= 0.5036 g / L

= 503.6 mg / L

= 503.6 ppm

in ppm Ca+2 concentration = 503.6 ppm

4 )

1) before titration starts

moles of benzoic acid = 0.2915 / 122.12 = 2.39 x 10^-3

molarity = moles / volume = 2.39 x 10^-3 / 100 x 10^-3 = 0.0239 M

C = 0.0239 , Ka = 6.3 x 10^-5

[H+] = sqrt (Ka x C) = sqrt (6.3 x 10^-5 x 0.0239) = 1.227 x 10^-3 M

pH = -log [H+] = -log (1.227 x 10^-3)

pH = 2.91

2) pH at half-equivalence point:

at this point . pH = pKa

pKa = -log Ka = -log (6.3 x 10^-5) = 4.20

pH = 4.20