25.0cm³ of hydrochloric acid is being neutralized by 20.0 cm³ of a solution of 0.15mlo dm-³ sodium carbonate solution. Q1).How many moles of sodium carbonate are neutralized by 1.o mol of HCl?
Q2).What is the concentration of the hydrochloric acid?
Q3). A tanker of acid is emptied into a water supply by mistake. A water company company christ titrates the water and finds 10.0dm³ of water are needed to neutralize 10.0dm-³ solution of sodium hydroxide. What is the concentration of hydrogen ions in the water?
Q1) For every 2 moles of hydrochloric acid, 1 mole of sodium carbonate is neutralized according to the balanced chemical equation:
2HCl + Na2CO3 → 2NaCl + H2O + CO2
Therefore, 1 mole of HCl will neutralize 0.5 moles of Na2CO3.
Q2) The concentration of the hydrochloric acid can be calculated using the formula:
Concentration = (Volume of solution (dm³) x molarity) / Volume of acid (dm³)
Given that 25.0 cm³ of HCl is neutralized by 20.0 cm³ of 0.15 mol dm⁻³ Na2CO3 solution:
Concentration = (20.0 cm³ x 0.15 mol dm⁻³) / 25.0 cm³
Concentration = 0.12 mol dm⁻³
Q3) In the neutralization reaction between sodium hydroxide and hydrochloric acid, we can see that 1 mole of HCl will neutralize 1 mole of NaOH:
HCl + NaOH → NaCl + H2O
Therefore, since 10.0 dm³ of sodium hydroxide solution is neutralized by 10.0 dm³ of water:
Concentration of hydrogen ions in water = 1 / 10.0 dm³
Concentration of hydrogen ions in water = 0.1 mol dm⁻³