Question

50.0 mL of 0.100 mol/L HCI is added to 50.0 mL of 1.00 mol/L NaOH Calculate the
hydronium and hydroxide concentrations o f the resulting solution. Explain.

Answer 1

Given that - Molarity of HCl = 0.100m Volume of Hel - 50.0 m2 Molarity of NaoH = 100m Volume of NaOH = 50.0m2 ho. of mole of Hcl = no. of mole of Hel = hoi of mole of NaoH = molarity of Hcl x Volume of HCl 0.100 x 50 = 5.0 mole. Molarity of Nao H x Volume of NaOH ho. of mole of NaoH = 1000 5000 = 50 mole Rid Write neutralization Reaction HCl + NaOH Nach +H2O Imole Imole Imole i mole According to neutralization reaction one mole of Hel Neutralizes I mole of Haoh to produce one mole of Nach s me mole of H2O so 5.0 mole of Hcl nu neutralizes the 5050 mole of Naoh So remaining NaOH= 50.0 mole - 500 mole = 45.0 mL Naoh so 45.0 mole NaOH remain in the solution. Noon - Nat toho mole Imole Imole as Imole of Naot give imole of OH in So usiomole of Naot give 45.0 mole of one in so the concentration of one in in resulting solution with will be us o mole