4. Write a chemical equation for a neutralization reaction between H2SO4 (strong

Question

4. Write a chemical equation for a neutralization reaction between H2SO4 (strong acid) and NH3 (weak base, Kb = 1.8 x 10-5).

a. If they both start as 0.1 M solutions, what are their starting pH values (before being mixed)?

b. Speculate on the strength of the conjugate acid of NH3 and calculate its Ka value. Before mixing, which substance has a higher concentration at equilibrium, NH3 or its conjugate acid?

c. What is the pH after mixing 150 ml of 0.1M H2SO4 and 100 ml of 0.1M NH3?

d. Does the Kb of NH3 matter in this problem? Why or why not?

Explanation / Answer

H2SO4 + NH3 -----------------> NH4 HSO4

a) 0.1 M H2SO4

the [H+] = 2 x0.1 M as it is a dibasic acid

hence pH = -log [H+]

= -log (0.2)

= 0.6989

0.1 M NH3

pH = 14 - (pOH

= 14 - 1/2[pKb -logC]

= 14 -1/2 [4.76 -log 0.1]

= 12.12

b) As NH3 is a weak base the conjugate acid NH4+ is relatively stronger.

We have the relation of base and its conjugate acid is

Ka x kb = Kw

thus Ka of conjugate acid of Nh3 is = 1.0x10-14 / 1.8x10-5

= 5.55x10-10

Before mixing the concnetration of NH3 is more than its conjugate acid.

Part C)

H2SO4 + NH3 -----------------> NH4+ + HSO4-

2 x 0.1x 150 100x0.1 0 0 initial mmoles

30 10

20 0 10 0 after reaction

Thus the solution has a strong acid and a weak conjugate acid.

due to common effect all the [H+ comes from strong acid only .

Thus [H+] = 20 /250= 0.08 M

henc epH = - log 0.08

=1.096

d) ther is no involvement of kb of NH3 inthis pH calculation , as the solution is acidic not a buffer.

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