5. 8.4 mg NaHCO3 powder was slowly added to 10 ml 0.03 M HCl in a 50ml beaker with
stirring. After all effervescence has seized, what should be the pH of the solution?
a. 1.69
b. 2.00
c. 2.23
8.4 mg NaHCO3 in 10 ml 0.03 M HCl
The reaction between NaHCO3 and HCl produces CO2 gas, H2O, and NaCl, with effervescence ceasing when NaHCO3 is fully consumed, leaving excess strong acid HCl.
Reaction Stoichiometry
NaHCO3 fully neutralizes with HCl via:
NaHCO3 + HCl → NaCl + H2O + CO2 ↑Moles of NaHCO3 =
8.4×10⁻³ g / 84 g/mol = 0.0001 molMoles of HCl =
0.03 M × 0.010 L = 0.0003 molExcess HCl = 0.0003 – 0.0001 = 0.0002 mol remains after reaction.
Final pH Determination
Total volume ≈ 10 mL (0.010 L), as solid volume is negligible.
[H⁺] = 0.0002 mol / 0.010 L = 0.020 MpH =
-log(0.020) = 1.70 (rounds to option a. 1.69)Option Analysis
- a. 1.69: Correct. Matches calculated excess [H⁺] of ~0.02 M for strong acid post-reaction.
- b. 2.00: Incorrect. Equals pH of original 0.03 M HCl (
-log(0.03)≈1.52), ignores partial neutralization. - c. 2.23: Incorrect. Assumes different excess (~0.006 M HCl,
-log(0.0059)≈2.23), but stoichiometry shows 0.02 M.
