QuestionIn a titration, if $15 \mathrm{~cm}^{3}$ of $0.15 \mathrm{M}$ HCl neutralizes $25 \mathrm{~cm}^{3}$ of sodium carbonate, find its concentration. (a) $0.025 \mathrm{M}$ (b) $0.045 \mathrm{M}$ (c) $0.0125 \mathrm{M}$ (d) $0.065 \mathrm{M}$

Studdy Solution
Calculate the concentration of the sodium carbonate solution.
$$Concentration\, of\, Na2CO =0.001125 \mathrm{~moles} /0.025 \mathrm{~} =0.045 \mathrm{M}$$The concentration of the sodium carbonate solution is $0.045 \mathrm{M}$.