Titration (Acid-Base Reactions) If a 17.5 mL sample of 1.3 M solution of each of
ID: 536961 • Letter: T
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
Explanation / Answer
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
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