Four solutions of unknown HCl concentration are titrated with solutions of NaOH. The following table lists the volume of each unknown HCl solution, the volume of NaOH solution required to reach the equivalence point, and the concentration of each NaOH solution. Calculate the concentration (in M) of the unknown HCl solution in each case.

HCl Volume (mL) |
NaOH Volume (mL) |
[NaOH] (M) |

(a) 25.00 Ml |
28.44 mL |
0.1231 M |

(b) 15.00 mL |
21.22 mL |
0.0972 M |

(c) 20.00 mL |
14.88 mL |
0.1178 M |

(d) 5.00 mL |
6.88 mL |
0.1325 M |

Solution: Here, we are going to calculate the concentration of the unknown HCl solution.

Step1:

Neutralisation reaction is the reaction between an acid and a base to give salt and water. The neutralization reaction in the given problem is:

NaOH(aq) + HCl(aq) →H2O(l) + NaCl(aq)

Here, 1 mole of HCl requires 1 mole of NaOH for complete neutralization.

Step2:

In titration calculations, the unknown concentration can be calculated by the following formula:

Where, VA = volume of acid used

VB = volume of base used

MA = concentration of acid

MB = concentration of base

nA = no. of moles of acid in balanced equation for reaction

nB = no. of moles of base in balanced equation for reaction

Step3:

Given, VA = 25.00 mLVB = 28.44 mL

MA = concentration of acid

MB = 0.1231 M

nA = 1

nB = 1

Substituting the values in the above equation, we get,

25 mL x MA = 28.44 mL x 0.1231 M

MA = 3.500964 / 25 M

MA = 0.14 M

Thus, concentration of the unknown HCl solution is 0.14 M.

Step4:...