Calculate the molarity of the HCl from the volumes of acid and base...?

at the equivalence point and the molarity of the NaOH.
Show your calculations.

HCl=25 ml (unknown concentration);
NaOH=15.0004 ml (1M/50 ml)

Answer:
you show 1M which means that NaOH is 1 molar (not sure what you mean by 1M/50 ml), From the definition of molar concentration C=n/V (C-conc., n= number of moles, V= volume), at the equivalence point n(NaOH)=n(HCl), so
15.0004ml*1M=25ml*x, x=(15.0004*1)/25=0.6 M (significant numbers note, since only one digit here). So the HCl solution is 0.6 M
HCl + NaOH ---> NaCl + H2O

The number of moles oh HCl is the same as the number of moles of NaOH:
n(HCl) = n(NaOH)

We know that c = n/V
which gives us
n = c × V

so:
c(HCl) × V(HCl) = c(NaOH) × V(NaOH)

therefore:
c(HCl) = c(NaOH) × V(NaOH) / V(HCl)

c(NaOH) = n/V = 1 mole / 0,05 L = 20 mol/L which is unreal...

c(HCl) = 20 mol/L × 0,0150004 L / 0,025 L
c(HCl) = 12,00032 mol/L which is also unreal...

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