Number of moles of NaOH = M(NaOH) *
V(NaOH)
                                                    Â
= 2 M * 0.08 L
                                                      Â
= 0.16 mol
Number of moles of HCl = M(HCl) * V(HCl)
                                                    Â
= 4 M * 0.02 L
                                                      Â
= 0.08 mol
HCL + NaOH -->NaCL + H2O
0.08 mol of each will react and neutralise each other
After Reaction :
Number of molesof NaOH remaining = 0.16 - 0.08 =
0.08 mol
Total volume= 80 mL+ 20mL =100 mL = 0.1 L
[OH-]=[NaOH]=Number of moles /Volume
                          Â
= 0.08 / 0.1
                          Â
= 0.8 M
Answer: final concentration of OH- ions= 0.8 M
