When 50.0mL of 0.100M NH3 (Kb=1.8x10^-5) solution is mixed with 20.0mL of of 0.250M HCl solution, what is the pH after reaction?
I solved the ka=5.56x10^-10 and I solved the moles of each nHCl and nNH3 and both have 0.005 moles. What do I do after this?
After determining the number of moles of HCl and NH3, you can use the concept of stoichiometry to calculate the amount of excess or remaining reactants. Since HCl and NH3 react in a 1:1 mole ratio, it can be concluded that 0.005 moles of each will react completely.
Now, let's calculate the concentration of NH3 and HCl in the final solution after the reaction.
The total volume of the final solution can be found by adding the initial volumes of NH3 and HCl: 50.0 mL + 20.0 mL = 70.0 mL = 0.070 L.
To calculate the concentration of NH3 in the final solution (after reaction), use the equation:
(concentration) = (moles) / (volume)
NH3 concentration = 0.005 mol / 0.070 L = 0.0714 M
Similarly, the concentration of HCl in the final solution is:
HCl concentration = 0.005 mol / 0.070 L = 0.0714 M
Since HCl is a strong acid, it dissociates completely into H+ ions. Therefore, the concentration of H+ in the final solution is 0.0714 M.
To calculate the pH of the solution, we need to determine the pOH first. The reaction between NH3 and HCl produces water and a salt. In this case, the salt is NH4Cl.
Since NH4Cl is a strong electrolyte, it completely dissociates in water, leaving NH4+ and Cl- ions. Since NH4+ can act as a weak acid, it will partially dissociate in water and release H+ ions.
The equation for the dissociation of NH4+ is:
NH4+ + H2O → NH3 + H3O+
The concentration of H+ ions from NH4+ can be calculated using the Ka value of NH4+ and the initial concentration of NH3.
Now, using the Kb value of NH3, you can calculate the concentration of hydroxide ions (OH-) in the final solution.
Kb = [NH3][OH-] / [NH4+]
Kb = 1.8x10^-5
[NH3] = 0.0714 M (concentration of NH3)
[OH-] = ?
[NH4+] = [H+], since the initial concentration of NH4+ is negligible compared to the concentration of NH4+ from NH3 dissociation.
Using the Kb expression, we can rearrange it to solve for [OH-]:
[OH-] = (Kb x [NH4+]) / [NH3]
Substituting the values, we have:
[OH-] = (1.8x10^-5) x (0.0714) / 0.0714
[OH-] = 1.8x10^-5 M
Finally, we can use the relationship between [H+] and [OH-] to calculate the pOH:
pOH = -log10 ([OH-])
pOH = -log10 (1.8x10^-5)
pOH ≈ 4.74
Since pH + pOH = 14, we can find the pH of the solution:
pH = 14 - pOH
pH = 14 - 4.74
pH ≈ 9.26
Therefore, the pH of the solution after the reaction is approximately 9.26.
If the reaction is mole for mole, then y0u have the salt at the end of the reaction. Hydrolyze the salt (react it with water) and go from there.
NH4Cl ==> NH4^+ | Cl^-
The NH4^+ hydrolyzes.
NH4^+ + HOH ==> NH3 + H3O^+
Set up and ICE chart and solve for H3O^+ and from there you can get pH.