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titration question:
How many moles of HCl must have been present in the 25mL of HCl solution in the 2 trials?

Trial 1: V of HCl= 25.00mL, V of NaOH used=29.50mL and M of NaOH=0.18M
Trial 2: V of HCl=25.00mL, V of NaOH used=28.50mL and M of NaOH=0.18M

  • chemistry -

    Do it this way.
    mols NaOH = M x L = ?
    mols HCl = mols NaOH.

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