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?

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Explanation / Answer

a. Volume of NaOH added = 28.21 mL = 0.02821 L

Concentration of NaOH added = 0.1095 M = 0.1095 miol/L

Total moles of NaOH added = 0.02821 L X 0.1095 miol/L = 0.003089 mol

b. Volume of HCl added = 2.51 mL = 0.00251 L

Concentration of HCl added = 0.1064 M = 0.1064 miol/L

Total moles of HCl added = 0.00251 L X 0.1064 miol/L = 0.000267 mol

c. Total moles of NaOH = Moles H+ from HCl + Moles H+ from sample

Moles H+ from solid acid = Total moles of NaOH - Moles H+ from HCl = 0.003089 mol- 0.000267 mol

Moles H+ from solid acid = 0.002822 mol

Moles NaOH that only reacted with the solid acid = moles of H+ from solid acid = 0.002822 mol

d. Equivalent mass of the acid = Mass of acid / Moles H+ from acid = 0.1270 g / 0.002822 mol = 45.0035 g/mol

e. It is not a diprotic acid as titration curves of diprotic acids have two equivalence points (Volume of NaOH) but here only one equivalence point (28.21 mL  of NaOH) is obtained. So, it is a monoprotic acid.

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