Answer: 1.45M
Explanation:
The definition of molarity is moles/liter. The neutralization of NaOH with HI is:
HI + NaOH = NaI + H2O
One mole of HI reacts with 1 mole of NaOH. We'll assume this is a titration reaction and that the 10.00ml sample of NaOH contains the same number of moles as the 56.75ml of 0.256M HI.
Moles HI: (0.256 moles/liter)*(0.05675 L) = 0.01453 moles HI
That means we muct have 0.01453 moles NaOH in 10.0ml of NaOH solution.
(0.01453 moles NaOH)/(0.010L) = 1.45 M
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Another approach is to use the relationship M1V1 = M2V2, which is useful for titrations (M is concentration and V is volume):
We want M2, so rearrange: M2 = M1V1/V2
M2 = (0.256M)*(56.75ml)/(10.0ml)
M2 = 1.45M
