Not sure how to do these ones. Please walk through the steps. Thank you! 1. When

Question

Not sure how to do these ones. Please walk through the steps. Thank you!

1. When titrating 50.0 mL of 0.10 M H2SO4 with 0.10 M NaOH, how many mL of NaOH will you have added to reach the first equivalence point?

2. A student completes a titration of an unknown diprotic acid. In this experiment, 0.79 g of the acid is dissolved in 250.0 mL of water. It requires 13.48 mL of 1.0 M NaOH to reach the second equivalence point. What is the molar mass of the acid?

3. Give the balanced chemical reaction for the titration of a generic diprotic acid, H2X, with potassium hydroxide.

Explanation / Answer

1) since H2SO4 is a diprotic acid

therefore, its normality = 2*its molarity

or, its number of equivalents = 2*its number of moles

similarly, NaOH is a monoacidic base,

therefore,

its normality = its molarity

or, its number of equivalents = its number of moles

now, number of moles = molarity*volume of solution in litres

thus, moles of H2SO4 = 0.1*0.05 = 0.005

number of equivalents of H2SO4 = 2*0.005 = 0.01

now, to reach the first equivalece point, equivalents of H2SO4 used = 0.01/2 = 0.005

thus, equivalents of NaOH required = 0.005 = moles of NaOH required

thus volume of NaOh required = moles /molarity = 0.005/0.1 = 0.05 litres = 50 ml

2) let the molar mass of the diprotic acid be 'M'

now, the equivalent mass of the acid will be 2M

now, moles of NaOH used = molarity*volume in litres = 1*0.0138 = 0.01348 moles

or, equivalents of NaOH used = 0.01348

thus equivalents of diprotic acid in 0.79 g of it = 0.01348

now, 0.01348 equivalents of the acid weighs 0.79 g

thus 1 equivalent of the acid weighs = 0.79/0.01348 = 58.61 g

or, 2M = 58.61 g

thus M = 117.21 g

3) H2A(aq) + KOH(aq) ------------> (HA-)(aq) + H2O(l) ..............first equivalence point

(HA-)(aq) + KOH(aq) ------------> (A2-)(aq) + H2O(l) .............second equivalence point

overall reaction :

H2A(aq) + KOH(aq) --------------> (A2-)(aq) + 2H2O(l)

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