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DirtyDon123
02-05-2011, 11:16 PM
This is a titration neutralisation question. I have 50cm3 of 0.1 M HCL and it is neutralised by 24cm3 NaOH. What is the concentration of the NaOH? Can you please show all steps in the working please. Many thanks and reps to all who help.

02-06-2011, 12:45 AM
First write out balanced equation....so cm3 is equal to one mL so convert hcl into number of moles by multiplying molarity by volume ( convert 50 ml to L first), after that, you use the mole to mole ratio to find number of moles of NaOH, then to find concentration just divide moles by volume again ( remember to convert to litres before dividing), this will give you molarity

forthenoodz
02-06-2011, 01:37 AM
0.208 3sf

Movjam
02-06-2011, 08:29 AM
This is a titration neutralisation question. I have 50cm3 of 0.1 M HCL and it is neutralised by 24cm3 NaOH. What is the concentration of the NaOH? Can you please show all steps in the working please. Many thanks and reps to all who help.
Reaction is:
NaOH + HCl ==> H2O + NaCl

Mole ratio is 1:1 thus equate their moles (moles = concentration x volume). Thus:

(50/1000) x 0.1 = (24/1000) x M
M being the concentration/molarity of NaOH. Solve for M, M = 0.208 to 3 sf.