Abeaker contains 47.0 ml of a 0.47 m hi solution and is titrated with 25.0 ml of a 0.25 m of koh. what is the final ph, and where is this point in the titration?

-

1.8

End point passed.

Explanation: -

Volume of HI solution = 47.0 mL = 0.047 L

Strength of HI solution = 0.47 M

Since HI is a strong acid, all of HI will dissociate to give H +.

[H+ ] =0.47 M x 0.047 L

= 0.02209 mol

Volume of KOH = 25.0 mL = 0.025 L

Strength of KOH = 0.25 M

Since KOH is a strong base, all of KOH will dissociate to give OH-.

[OH-] = 0.25 M x 0.025L

= 0.00625 mol

Since [H+] and [OH-] react to form water,

[H+] unreacted = 0.02209 – 0.00625 = 0.01584 mol

Using the formula

pH = - log [H+]

= - log 0.01584

= 1.8

As the strong acid HI is being titrated by strong base KOH, the pH at the end point should be 7.

The pH has already crossed that. Thus the titration end point has already passed

a gas, having a molecular weight of 13 kg/kmol and a kinematic viscosity of 0.25 cm2/s, flows through a pipe 0.25 m internal diameter and 5 km long at the rate of 0.4 m3/s and is delivered at atmospheric pressure.