A250. ml sample of 0.0328m hcl is partially neutralized by the addition of 100. ml of 0.0245m naoh. find the concentration of hydrochloric acid in the resulting solution.

0.0164 molar concentration of hydrochloric acid in the resulting solution.

Explanation:

1) Molarity of 0.250 L HCl solution : 0.0328 M

Moles of HCl in 0.250 L solution = 0.0082 moles

2) Molarity of 0.100 L NaOH solution : 0.0245 M

Moles of NaOH in 0.100 L solution = 0.00245 moles

3) Concentration of hydrochloric acid in the resulting solution.

0.00245 moles of NaOH will neutralize 0.00245 moles of HCl out of 0.0082 moles of HCl.

Now the new volume of the solution = 0.100 L +0.250 L = 0.350 L

Moles of HCl left un-neutralized = 0.0082 moles - 0.00245 moles =  0.00575 moles

Molarity of HCl left un-neutralized :

0.0164 molar concentration of hydrochloric acid in the resulting solution.

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