Question

Titration (Acid-Base Reactions)

If a 17.5 mL sample of 1.3 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) 17.5 mL of H₂SO₄ titrated with 0.90 M NaOH

-volume of NaOH? mL

-total volume? mL

(b) 17.5 mL of HBr titrated with 0.90 M NaOH

-volume of NaOH? mL

total volume? mL

(c) 17.5 mL of HF titrated with 0.90 M NaOH

-volume of NaOH? mL

-total volume? mL

Answer 1

a) 17.5 ml H2SO4( 1.3M ) is titrated with 0.9M NaOH

H2SO4 + 2 NaOH -------> Na2SO4 + 2H2O

Moles of H2SO4 = 0.0175 * 1.3 = 0.02275

The ratio between H2SO4 and NaOH is 1:2

Moles of NaOH needed = 2 * 0.02275 = 0.0455

Volume of NaOH needed = 0.0455/0.9 = 0.0505 L = 50.5 ml

Total volume at equivalence point = 17.5+50.5 ml = 68 ml

b) 17.5 ml HBr (1.3M) titrated with 0.9 M NaOH

HBr + NaOH -------> NaBr + H2O

Moles of HBr = 0.0175 * 1.3 = 0.02275

Ratio of HBr and NaOH is 1:1

Moles of NaOH required = 0.02275

Volume of NaOH required = 0.02275/0.9 = 0.02527 L = 25.27 ml

Total volume at equivalence point = 17.5 ml + 25.27 ml = 42.77 ml

c) the above same calculation refers to HF

17.5 ml HF (1.3M) titrated with 0.9 M NaOH

HF + NaOH ----> NaF + H2O

Moles of HF = 0.0175 * 1.3 = 0.02275

Ratio of HF and NaOH is 1:1

Moles of NaOH required = 0.02275

Volume of NaOH required = 0.02275/0.9 = 0.02527 L = 25.27 ml

Total volume at equivalence point = 17.5 ml + 25.27 ml= 42.77 ml