Question

The alkalinity of natural waters is usually controlled by OH- , CO3 2- , and HCO3 - , which may be present singularly or in combination. Titrating a 100.0 mL sample to a pH of 8.3 requires 25.0 mL of a 0.025 M solution of HCl. A second 100.0 mL aliquot requires 50.0 mL of the same titrant to reach pH of 4.5. Calculate the concentrations of CO3 2- and HCO3 - in ppm.

Answer 1

• CO₃⁻² ⇒ 375 ppm

• HCO₃⁻ ⇒ 381.25 ppm

Step-by-step explanation:

When titrating to a pH=8.3, the CO₃⁻² species are converted to HCO₃⁻, while a titration to a pH=4.5 both the CO₃⁻² and the HCO₃⁻ species are neutralized.

Using the data of the first titration we can calculate [CO₃⁻²] in the sample:

• 0.025 M * 0.025 L = [CO₃⁻²] * 0.100 L

• [CO₃⁻²] = 6.25x10⁻³ M

Using the data of the second titration we can calculate | [CO₃⁻²] + [HCO₃⁻] | in the sample:

• 0.025 M * 0.050 L = ( [CO₃⁻²] + [HCO₃⁻] ) * 0.100 L

• [CO₃⁻²] + [HCO₃⁻] = 0.0125 M

Using the data from the first step, we can now calculate [HCO₃⁻]:

• [HCO₃⁻] = 0.0125 M - [CO₃⁻²]

• [HCO₃⁻] = 6.25x10⁻³ M

Now we just need to convert the concentrations from M to ppm (mg/L)

• CO₃⁻² ⇒ 6.25x10⁻³
  `(mol)/(L)*(60g)/(1mol)*(1000mg)/(1g)=` 375 ppm

• HCO₃⁻ ⇒ 6.25x10⁻³
  `(mol)/(L)*(61g)/(1mol)*(1000mg)/(1g)=` 381.25 ppm