Answer:
Concentration of the original [tex]\rm KOH[/tex] solution: approximately [tex]1.05\; \rm mol \cdot dm^{-3}[/tex].
Explanation:
Notice that the concentration of the [tex]\rm HCl[/tex] solution is in the unit [tex]\rm mol\cdot dm^{-3}[/tex]. However, the unit of the two volumes is [tex]\rm cm^{3}[/tex]. Convert the unit of the two volumes to [tex]\rm dm^{3}[/tex] to match the unit of concentration.
[tex]\begin{aligned} V(\mathrm{NaOH}) &= 25.0\; \rm cm^{3} \\ &= 25.0\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{1000\; \rm cm^{3}} \approx 0.0250\; \rm dm^{3} \end{aligned}[/tex].
[tex]\begin{aligned} V(\mathrm{HCl}) &= 35.0\; \rm cm^{3} \\ &= 35.0\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{1000\; \rm cm^{3}} \approx 0.0350\; \rm dm^{3} \end{aligned}[/tex].
Calculate the number of moles of [tex]\rm HCl[/tex] molecules in that [tex]0.0350\; \rm dm^{3}[/tex] of [tex]0.75\; \rm mol \cdot dm^{3}[/tex] [tex]\rm HCl\![/tex] solution:
[tex]\begin{aligned}n(\mathrm{HCl}) &= c(\mathrm{HCl}) \cdot V(\mathrm{HCl}) \\ &= 0.00350\; \rm dm^{3} \times 0.75\; \rm mol \cdot dm^{3} \\ &\approx 0.02625\; \rm mol\end{aligned}[/tex].
[tex]\rm HCl[/tex] is a monoprotic acid. In other words, each [tex]\rm HCl\![/tex] would release up to one proton [tex]\rm H^{+}[/tex].
On the other hand, [tex]\rm KOH[/tex] is a monoprotic base. Each [tex]\rm KOH\![/tex] formula unit would react with up to one [tex]\rm H^{+}[/tex].
Hence, [tex]\rm HCl[/tex] molecules and [tex]\rm KOH\![/tex] formula units would react at a one-to-one ratio.
[tex]{\rm HCl}\, (aq) + {\rm NaOH}\, (aq) \to {\rm NaCl}\, (aq) + {\rm H_2O}\, (l)[/tex].
Therefore, that [tex]0.02625\; \rm mol[/tex] of [tex]\rm HCl[/tex] molecules would neutralize exactly the same number of [tex]\rm NaOH[/tex] formula units. That is: [tex]n(\mathrm{NaOH}) = 0.02625\; \rm mol[/tex].
Calculate the concentration of a [tex]\rm NaOH[/tex] solution where [tex]V(\mathrm{NaOH}) = 0.0250\; \rm dm^{3}[/tex] and [tex]n(\mathrm{NaOH}) = 0.02625\; \rm mol[/tex]:
[tex]\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} \\ &= \frac{0.02625\; \rm mol}{0.0250\; \rm dm^{3}}\approx 1.05\; \rm mol \cdot dm^{-3}\end{aligned}[/tex].
