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Description

Calculate the volume in mL of a 0.117 M NaOH solution required to titrate

10.00 mL of 0.135 M solution HCl(aq) solution. (units in mL)
10.00 mL of 0.12 M solution H2SO4(aq) solution. (units in mL)

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Explanation & Answer

In 10ml of 0.135 M HCl, the mole of H+=0.135x10.00x10^-3=1.35x10^-3 mole.

So 1.35x10^-3 moles of OH- is needed to neutralize the H+ in HCl.

Therefore the volume of NaOH=1.35x10^-3/0.117=11.5x10^-3 L=11.5 ml


In 10.00 mL of 0.12 M H2SO4, the mole of H+=2x0.12x10.00x10^-3=2.4x10^-3 mole

So 2.4x10^-3 moles of OH- is needed to neutralize the H+ in H2SO4.

Therefore the volume of NaOH=2.4x10^-3/0.117=20.5x10^-3 L=20.5 ml


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