Consider the titration of 120.0 mL of 0.100 M HI by 0.250 M NaOH.
Part 1
Calculate the pH after 35.0 mL of NaOH has been added.
pH =
Part 2
What volume of NaOH must be added so the pH = 7.00?
_______mL NaOH
Number of moles of HI = 120*0.1 = 12 mmol
number of moles of NaOH = 35 * 0.25 = 8.75 mmol
NaOH + HI ------> NaI + H2O
remaining number of moles HI remainig = 12 - 8.75 = 3.25 mmol
concentration of HI = 3.25/(120+35) = 0.021 mol/L
pH = -log[H+]
pH = -log(0.021)
pH = 1.678
when, number of moles of H+ = number of moles OH- then pH of the solution = 7
we have the formula M1V1 = M2V2
0.1*120 = 0.25*V2
V2 = 48 mL
so if we add 48 mL of NaoH then the pH will become pH = 7
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