A 250. ml sample of 0.0328m hcl is partially neutralized by the addition of 100. ml of 0.0245m naoh. find the concentration of h

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ydrochloric acid in the resulting solution.

1 answer:

lions [2.9K] · Answer 1 · 2026-03-11T10:07:59+03:00

Answer: 0.0164 molar concentration of hydrochloric acid in the resulting solution.

Explanation:

1) Molarity of 0.250 L HCl solution: 0.0328 M

$Molarity=\frac{\text{Number of moles}}{\text{Volume of solution in liters}}=0.0328=\frac{\text{Number of moles}}{0.250 L}$

The amount of HCl in the 0.250 L solution = 0.0082 moles

2) Molarity of 0.100 L NaOH solution: 0.0245 M

$Molarity=\frac{\text{Number of moles}}{\text{Volume of solution in liters}}=0.0245=\frac{\text{Number of moles}}{0.100 L}$

The amount of NaOH in the 0.100 L solution = 0.00245 moles

3) Determining the concentration of hydrochloric acid in the final solution.

0.00245 moles of NaOH will neutralize 0.00245 moles of HCl from the original 0.0082 moles of HCl.

The total volume of the mixture becomes 0.100 L + 0.250 L = 0.350 L

Remaining moles of unreacted HCl = 0.0082 moles - 0.00245 moles = 0.00575 moles

$Molarity=\frac{\text{number of moles}}{\text{volume of solution in L}}$

Concentration of the remaining HCl: $\frac{0.00575 moles}{0.350L}=0.0164 M$

0.0164 molar concentration of hydrochloric acid in the resulting solution.