Q.57 The concentration of NaCl (in mM) formed at the stoichiometric equivalence point when 10 mL of 0.1 M HCl solution is titrated with 0.2 M NaOH solution is ________. (as an integer)
The concentration of NaCl formed at the stoichiometric equivalence point is 67 mM. This integer value accounts for the total moles of NaCl produced divided by the final solution volume, rounded appropriately for the given precision. Calculations confirm this through standard acid-base titration stoichiometry.
Titration Setup
HCl reacts completely with NaOH in a 1:1 molar ratio: HCl + NaOH → NaCl + H2O. For 10 mL (0.010 L) of 0.1 M HCl, moles of HCl equal 0.001. Equivalence requires 5 mL (0.005 L) of 0.2 M NaOH, as moles NaOH = moles HCl.
Concentration Calculation
Total volume at equivalence is 15 mL (0.015 L). NaCl concentration is 0.001 moles / 0.015 L = 0.0667 M, or 66.7 mM, reported as integer 67. No options provided in query; common errors include ignoring volume addition (100 mM) or wrong stoichiometry.
Step-by-Step Solution
Moles of HCl: 0.1 M × 0.010 L = 0.001 mol
NaOH volume: V = 0.001 / 0.2 = 0.005 L = 5 mL
Total volume: 10 mL + 5 mL = 15 mL = 0.015 L
NaCl [M]: 0.001 / 0.015 = 0.0667 M
NaCl [mM]: 66.7 ≈ 67 (integer)
Common Mistakes Explained
- Forgetting titrant volume yields 100 mM (wrong).
- Miscalculating moles gives 50 mM or 33 mM.
- No dilution adjustment ignores total volume effect.
CSIR NET Relevance
This tests stoichiometry in strong acid-strong base titrations, key for quantitative analysis sections. Practice similar problems for precision.
