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1.) 10 mL of 0.1 M NaOH is added to 50 mL of 0.1 M of the hypothetical acid HA.

ID: 738585 • Letter: 1

Question

1.) 10 mL of 0.1 M NaOH is added to 50 mL of 0.1 M of the hypothetical acid HA. write the chemical equation and calculate the molarity of NaA and HA in the final solution.
Please show your work.

Explanation / Answer

NaOH + HA = NaA + H2O x/0.01 = 0.1 =>x = 0.001....No. of moles of NaOH y/0.05 = 0.1 =>y = 0.005....No. of moles of HA Number of moles of NaA formed = 0.001 Number of moles of HA remaining = 0.004 Number of moles H20 formed = 0.001 Total volume = 10+50 = 60mL = 0.06 Therefore molarity of NaA = 0.001/0.06 = 0.01667 and molarity of HA = 0.004/0.06 = 0.0667

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