Question

An environmental chemist obtained a 200 mL sample of lake water believed to be contaminated with a single monoprotic strong acid. Titrating this sample with a 0.0050 M NaOH(aq) required 7.3 mL of the NaOH solution to reach the endpoint. If the size of the lake can be approximated as 1.1 km long by 2.3 km wide, and has an average depth of 10 m, estimate how many moles of the strong acid are present in the lake?

Answer 1

The chemical reaction between the monoprotic acid and NaOH is:

                              HA + NaOH ---------------------> NaA + H2O

Hence one mole of acid is reacting with one mole of base.

No. of moles = volume in Litres * molarity

                         HA                      +               NaOH ---------------------> NaA + H2O

                    0.2 L *x M                            0.005M * 0.0073L

    We know that: equal moles of acid and base(NaOH) are reacting from the above balanced equation.

Hence             0.2 L *x M   = 0.005M * 0.0073L

x = (0.005 M * 0.0073L)/ 0.2L

        = 0.0001825 M

Hence molarity of monoprotic acid in 200ml water sample is---------------- 0.0001825M

Volume of water in the lake = 1.1 *10^3 m * 2.3 *10^3m * 10 m

                                                 = 25.3*10^6 m³

                                                 = 25.3 *10^9 litres                                 [ since 1m3 = 1000litres]

No.of moles of acid= molarity* volume in L= 0.0001825M* 25.3 *10^9 litres   

                                                                            = 4.6* 10^6 moles