The student dissolved 2.570 grams of acid in 50 mL of distilled water. He then t

Question

The student dissolved 2.570 grams of acid in 50 mL of distilled water. He then titrated the unknown acid with a standardized NaOH solution. Unfortunately, the student over shot the end point and had to back titrate the solution with 3.50 mL of 0.99897 M HCl. Once the final end point was determined, the student has added a total of 33.55 mL of a 1.0446 M NaOH solution. How many moles of OH were used to reach the end point? How many moles of H from HCl were used for back titration? How many moles of H from the solid were neutralized? What is the equivalent mass of the unknown acid?

Explanation / Answer

moles of OH were used to reach the end point = 0.03355 x 1.0446 = 0.03504 moles of OH-

moles of OH were used to reach the end point = 0.0035 X 0.99897 = 0.0034964 moles

Moles of H+ from solid neutralized = 0.03504 - 0.0034964 = 0.03155 moles

Equivalent mass = 2.57 / 0.03155 = 81.46 grams

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