While working in the laboratory, a technician accidentally knocks over a beaker and spills all 200.0 mL of 2.5 M HCl.
Which of the following would neutralize the spill and not have any base left over?

A) 200.0 mL of 2.5 M Ca(OH)2
B) 100.0 mL of 5.0 M KOH
C) 500.0 mL of 1.0 M NaOH
D) 125.0 mL of 2.0 MCa(OH)2
E) 250.0 mL of 2.0 MCa(OH)2

500.0 mL of 1.0 M NaOH

Explanation:

The number of moles of acid contained in the spill is given by

n=CV

C= concentration of acid= 2.5 M

V= volume of acid = 200mL

n= 2.5 × 200/1000 = 0.5 moles

Hence 0.5 moles of base is required for complete neutralization. Looking at the options, 500.0 mL of 1.0 M NaOH contains 0.5 moles of NaOH.

The reaction is in 1:1 molar ratio.

NaOH(aq) + HCl(aq) > NaCl(aq) + H2O(l)

Hence 0.5 moles of HCl will react with 0.5 moles of NaOH and exactly neutralize the spilt acid.

it is stable as it is in group 0 which have full outer shells of electrons