A sample of ammonia (Kb = 1.8 * 10-5) is titrated with 0.1 M H2SO4. At the equivalence point, the pH is approximately
Isn't it should be 5 right i believe.
Approximately 5 is ok.
To determine the pH at the equivalence point of this titration, we need to know the balanced chemical equation for the reaction between ammonia (NH3) and sulfuric acid (H2SO4). The balanced equation is as follows:
2 NH3 + H2SO4 -> (NH4)2SO4
From this equation, we can see that for every 2 moles of ammonia, 1 mole of sulfuric acid is required to reach the equivalence point.
At the equivalence point, all of the ammonia has reacted with the sulfuric acid, resulting in the formation of the ammonium sulfate salt, (NH4)2SO4.
(NH4)2SO4 is a salt that completely dissociates in water to produce ammonium ions (NH4+) and sulfate ions (SO4^2-). The ammonium ion is the conjugate acid of ammonia, which means it can also act as a weak acid in water.
To determine the pH at the equivalence point, we need to calculate the concentration of the ammonium ion, NH4+, which can be found using the stoichiometry of the reaction.
In this case, since the concentration of the H2SO4 is 0.1 M, and the reaction requires 2 moles of NH3 to neutralize 1 mole of H2SO4, we have:
(0.1 M H2SO4) / 2 = 0.05 M NH3 (since the ratio of NH3 to H2SO4 is 2:1)
Now, we can use the Kb value of ammonia to calculate the concentration of hydroxide ions (OH-) formed:
Kb = [NH4+][OH-] / [NH3]
Rearranging the equation, we have:
[OH-] = (Kb * [NH3]) / [NH4+]
= (1.8 * 10^-5) * (0.05 M) / (0.05 M)
= 1.8 * 10^-5 M
Since the concentration of hydroxide ions at the equivalence point is known, we can calculate the pOH:
pOH = -log10([OH-])
= -log10(1.8 * 10^-5)
= 4.74
Finally, we can calculate the pH:
pH = 14 - pOH
= 14 - 4.74
= 9.26
Therefore, at the equivalence point of the titration, the pH is approximately 9.26.