Question

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

​

Answer 1

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