A 0.1270 g sample of solid acid is titrated with 28.21 mL of 0.1095 M NaOH. This

Question

A 0.1270 g sample of solid acid is titrated with 28.21 mL of 0.1095 M NaOH. This overshot the endpoint and 2.51 mL of 0.1064 M HCl was used in the back titration. (Back Titration is when you add excess NaOH and have to use HCl to neutralize the excess) Total moles of NaOH = (moles H+ from HCl) + (moles H+ from sample)

a. Calculate the total moles of NaOH added

b. Calculate the total moles of HCl added

c. Calculate the moles NaOH that only reacted with the solid acid (This also equals the moles of H+ from the solid acid).

d. Calculate the equivalent mass of the acid (grams acid per mole H+).

e. If this is a diprotic acid, what is the molar mass?

Explanation / Answer

a. Calculate the total moles of NaOH added

Ans: 0.00308 moles of NaOH

b. Calculate the total moles of HCl added

Ans: 0.00026 moles of HCl

c. Calculate the moles NaOH that only reacted with the solid acid (This also equals the moles of H+ from the solid acid).

Ans: 0.00282 moles of Unknown acid

d. Calculate the equivalent mass of the acid (grams acid per mole H+).

Ans: 45g/mol

e. If this is a diprotic acid, what is the molar mass?

22.5g/mol

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