Question

If a 16.5 mL sample of 1.4 M solution of each of the following acids is reacted with 0.90 M NaOH, how many milliliters of the NaOH are required for the titration? What is the total volume (in mL) of solution at the equivalence point? (a) 16.5 mL of H,So, titrated with 0.90 M NaOH volume of NaOH mL total volume mL (b) 16.5 mL of HCl titrated with 0.90 M NaOH volume of NaOH ml total volume ml (c) 16.5 mL of HClO titrated with 0.90 M NaOH volume of NaOH ml total volume

Answer 1

a) Consider a reaction, H₂SO₄ + 2 NaOH Na₂SO₄ + 2 H₂O

From reaction, stoichiometric ratio = No of moles of acid / No of moles of base = 1/2

We have correlation, M _acid$\times$ V _acid = M _base$\times$ V _base$\times$ stoichiometric ratio.

Therefore, 1.4 M  $\times$ 16.5 ml = 0.90 M $\times$ V _base$\times$ (1/2)

V _base = 1.4 M  $\times$ 16.5 ml / 0.90 M $\times$ (1/2)

V _base = 51.3 ml

ANSWER: Volume of NaOH solution at equivalence point = 51.3 ml

Total volume of solution at equivalence point = Volume of HCl solution + Volume of added NaOH solution

Total volume of solution at equivalence point = 16.5 ml + 51.3 ml = 67.8 ml

b) Consider a reaction, HCl + NaOH NaCl + H₂O

From reaction, stoichiometric ratio = No of moles of acid / No of moles of base

stoichiometric ratio = No. of moles of HCl / No. of moles of NaOH = 1/1=1

We have correlation, M _acid$\times$ V _acid = M _base$\times$ V _base$\times$ stoichiometric ratio.

Therefore, 1.4 M  $\times$ 16.5 ml = 0.90 M $\times$ V _base$\times$ 1

V _base = 1.4 M $\times$ 16.5 ml / 0.90 M $\times$ 1

Volume of NaOH at equivalence point = 25.7 ml

Total volume of solution at equivalence point = Volume of HCl solution + Volume of added NaOH solution

Total volume of solution at equivalence point = 16.5 ml + 25.7 ml = 42.2 ml

ANSWER : Volume of NaOH = 25.7 ml

Total volume of solution at equivalence point = 42.7 ml

C)

Consider a reaction, HClO+ NaOH NaClO + H₂O

From reaction, stoichiometric ratio = No of moles of acid / No of moles of base

stoichiometric ratio = No. of moles of HClO / No. of moles of NaOH = 1/1=1

We have correlation, M _acid$\times$ V _acid = M _base$\times$ V _base$\times$ stoichiometric ratio.

Therefore, 1.4 M  $\times$ 16.5 ml = 0.90 M $\times$ V _base$\times$ 1

V _base = 1.4 M $\times$ 16.5 ml / 0.90 M $\times$ 1

Volume of NaOH at equivalence point = 25.7 ml

Total volume of solution at equivalence point = Volume of HClO solution + Volume of added NaOH solution

Total volume of solution at equivalence point = 16.5 ml +25.7 ml = 42.2 ml

ANSWER : Volume of NaOH at equivalence point = 25.7 ml

Total volume of solution at equivalence point = 42.2 ml