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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?

You have 25.0mL HCl of unknown concentration. It takes 12.5 mL of 0.2 M NaOH to neutralize-example-1
User Brad Davis
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1 Answer

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Answer:

1) The concentration of HCl = 0.1 M.

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

Step-by-step explanation:

1) The concentration of HCl:

  • 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: (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.

User Jzadeh
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