Question

A 26.9 mL sample of 0.349 M ammonia, NH₃, is titrated with 0.262 M hydroiodic acid.

At the equivalence point, the pH is

.

Use the Tables link in the References for any equilibrium constants that are required.

Answer 1

find the volume of HI used to reach equivalence point

M(NH3)*V(NH3) =M(HI)*V(HI)

0.349 M *26.9 mL = 0.262M *V(HI)

V(HI) = 35.8324 mL

Given:

M(HI) = 0.262 M

V(HI) = 35.8324 mL

M(NH3) = 0.349 M

V(NH3) = 26.9 mL

mol(HI) = M(HI) * V(HI)

mol(HI) = 0.262 M * 35.8324 mL = 9.3881 mmol

mol(NH3) = M(NH3) * V(NH3)

mol(NH3) = 0.349 M * 26.9 mL = 9.3881 mmol

We have:

mol(HI) = 9.3881 mmol

mol(NH3) = 9.3881 mmol

9.3881 mmol of both will react to form NH4+ and H2O

NH4+ here is strong acid

NH4+ formed = 9.3881 mmol

Volume of Solution = 35.8324 + 26.9 = 62.7324 mL

Ka of NH4+ = Kw/Kb = 1.0E-14/1.8E-5 = 5.556*10^-10

concentration ofNH4+,c = 9.3881 mmol/62.7324 mL = 0.1497 M

NH4+ + H2O -----> NH3 + H+

0.1497 0 0

0.1497-x x x

Ka = [H+][NH3]/[NH4+]

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((5.556*10^-10)*0.1497) = 9.118*10^-6

since c is much greater than x, our assumption is correct

so, x = 9.118*10^-6 M

[H+] = x = 9.118*10^-6 M

use:

pH = -log [H+]

= -log (9.118*10^-6)

= 5.0401

Answer: 5.04