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    7.9 ml of HCl neutralized 10 ml of 0.1 M NaOH.
    Would you say 0.001 moles HCl are reacting with 0.001 moles of NaOH since
    the reaction is 1:1 or would it be 0.00079 moles HCl reacting with 0.001 moles NaOH?
    Please help.
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    0.001 moles HCl are reacting with 0.001 moles of NaOH
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    The question asks to "Calculate how many moles of HCl are present in the average volume at the end-point"
    So there are 0.001 moles in 7.9 ml? And then the molarity would be 0.001/0.0079 L = 0.126 M. Less HCl was needed because its molarity was higher than the NaOH. Is that right?
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    Can anyone help me here please?
 
 
 
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