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By titration, it is found that 50.1 mL of 0.153 M NaOH(aq) is needed to neutralize 25.0 mL of HCl(aq). Calculate theconcentration of the HCl solution.[HCI) =M

By titration it is found that 501 mL of 0153 M NaOHaq is needed to neutralize 250 mL of HClaq Calculate theconcentration of the HCl solutionHCI M class=

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Explanation:

We are tritating a 25.0 mL solution of HCl with a solution of NaOH.

NaOH + HCl ----> NaCl + H₂O

We found that after adding 50.1 mL of a 0.153 M solution of NaOH we neutralized all the HCl. We reached the equivalence point. In that point we have the same number of equivalents of both species.

EQNaOH = EQHCl

Since their molar ratio is 1 to 1 (because both coefficients are 1) we can find the number of equivalents by multiplying the volume by the molarity.

EQNaOH = EQHCl

VNaOH * MNaOH = VHCl * MHCl

We already know the volume of both solutions and the concentration of the NaOH solution. We can solve that equation to get the molarity of the HCl solution.

VNaOH * MNaOH = VHCl * MHCl

50.1 mL * 0.153 M = 25.0 mL * MHCl

MHCl = 50.1 mL * 0.153 M/(25.0 mL)

MHCl = 0.307 M

Answer: the concentration of the HCl solution is HCl.

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