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A 25 mL aliquot of an HCl solution is titrated with 0.100 M NaOH. The equivalence...

A 25 mL aliquot of an HCl solution is titrated with 0.100 M NaOH. The equivalence point is reached after 21.27 mL of the base were added. Calculate the concentration of the acid in the original solution, the pH of the original HCl solution and the original NaOH solution

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Answer #1

first find out the no ofmoles of Base using molarity and volume

no ofmoles of NaOH = Molarity of NaOH and voleme of NaOH in liters

Molarity of NaOH = 0.1M,

volume of NaOH = 21.27mL convert in to liters = 0.02127L

no ofmoles of NaOH = 0.1M x 0.02127 L = 0.002127 moles

moles of NaOH = 0.002127 moles

write balanced equation

NaOH + HCl ---> NaCl + H2O

from this balanced equation it is clear that onemole of HCl required one mole of NaOH to reach equivelence point

0.002127 moles of NaOH required 0.002127 moles of HCl

now we have moles of HCl and volume of HCl we can find out the concentration

volume of HCl = 25mL convert in to liters = 0.025L

MOlarity of HCl = 0.002127 / 0.025L

= 0.085M

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