e lead peroxide to form water. When all the sulfuric acid is used up, the battery is "discharged" produces no current. The battery can be recharged by passing the current through it in the opposite direction. This process reverses all the previous reactions and forms lead at the anode and lead peroxide at the cathode. Proposed Problem i) The concentration of sulfuric acid is 0.0443 mol/L. The pH is: No. mol of hydrogen ions = 0.0443 mol/L x 2 = 0.0886 mol/L hydrogen ions pH = - log [H] = - log (0.0886) = - (-1.0525) = 1.05 Therefore, pH is 1.05. ii) The amount of base needed to neutralize the lake water is: volume of lake = 2000m x 800m x 50m = 800,000,000 m3 or 8x108 m3 since 1m3=1000L, therefore 8x1011 L 0.0443 mol/L x 8x1011 = 3.54 x 1010 mol of H2SO4 in water # mol NaOH = 3.54 x 1010 mol H2SO4 x 2 mol NaOH 1 mol H2SO4 = 7.08 x 1010 mol of NaOH needed Mass of NaOH = 7.08 x 1010 mol NaOH x 40 g NaOH 1 mol NaOH = 2.83 x 1012 g NaOH or 2.83 x 109 kg NaOH Therefore a total of 2.83 x 1012 g of NaOH is needed to neutralize the lake water. iii) The use of sodium hydroxide versus limestone to neutralize the lake water: Sodium hydroxide: Sodium hydroxide produces water when reacting with an acid, it also dissolves in water quite readily. When using sodium hydroxide to neutralize a lake, there may be several problems. One problem is that when sodium hydroxide dissolves in water, it gives off heat and this may harm aquatic living organisms. Besides this, vast amounts of sodium hydroxide is required to neutralize a lake therefore large amounts of this substance which is corrosive will have to be transported. This is a great risk to the environment if a spill was to occur. The following equation shows that water is produced when using sodium hydroxide. 2NaOH + H2SO4 * Na2 SO4 + 2H2O Limestone: Another way to neutralize a lake is by liming. Liming of lakes must be done with considerable caution and with an awareness that the aquatic ecosystem will...
