what volume of 2.5 NaOH is needed to neutralize a 135 ml of 1.15 M HCL solution?
We know that one mole of NaOH is required to neutralise one mole of HCl.
I.e 1 ml of 1M NaOH = 1 ml of 1M HCl
x ml of 2.5 M NaOH = 135 ml of 1.15 M HCl
Therefore x = 135*1.15/2.5
= 62.1 ml
Therefore 62.1 ml of NaOH is required.
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