Answer:
1) The concentration of HCl = 0.1 M.
2) The table that can be used to organize the information correctly is C.
Explanation:
<u><em>1) The concentration of HCl:</em></u>
- We know that the no. of millimoles of the acid is equal to the no. of millimoles of the base at the neutralization point.
which means that: <em>(MV)HCl = (MV)NaOH,</em>
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.
<em>2) The table that can be used to organize the information correctly is C</em>
<em></em>
<em>Table A and B are the same and reported volume of HCl and NaOH is wrong.</em>
<em>Table C is right, contain the correct volumes and concentration of NaOH and missed the concentration of HCl which is 0.1 M.</em>
<em>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.</em>