You have 25.0ml hcl of unknown concentration. it takes 12.5 ml of 0.2 m naoh to neutralize the acid. determine the concentration of hcl in the following steps.

which data table would you use to organize the information correctly?

1) The concentration of HCl = 0.1 M.

2) The table that can be used to organize the information correctly is C.

Explanation:

1) The concentration of HCl:

which means that: (MV)HCl = (MV)NaOH,

M of HCl = ??? M, V of HCl = 25.0 mL.

M of NaOH = 0.2 M, V of NaOH = 12.5 mL.

∴ M of HCl = (MV)NaOH/V of HCl = (0.2 M)(12.5 mL)/(25.0 mL) = 0.1 M.

2) The table that can be used to organize the information correctly is C

Table A and B are the same and reported volume of HCl and NaOH is wrong.

Table C is right, contain the correct volumes and concentration of NaOH and missed the concentration of HCl which is 0.1 M.

Table D reported the volume and the concentration of HCl wrongly and also the concentration of NaOH. The data reported of HCl and NaOH is reversed.

it forms an anion that has a larger radius. anions have a negative charge and are larger than the original atom.

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