1). In a titration 20 mL of 0.100 M of HCl is added to 50.0 mL of 0.150 M of ammonia.

A). What is the pH of NH3?

B). What is the pH of the resulting solution?

HELP :(

1) Makes no sense to me. Is that before the HCl is added or after. If after it's the pH of the solution, not NH3.

2)
HCl + NH3 ==> NH4Cl
mols HCl = M x L = 0.002
mols NH3 = M x L = 0.0075.
So you have how much NH3 that doesn't react? That's 0.0075 - 0.002 = 0.0055 mols NH3 excess. That's in how much volume? That's 50 mL + 20 mL = 70 mL = 0.07L so the excess NH3 concn is
mols/L = 0.0055/0.07 = 0.786M.
.......NH3 + H2O ==> NH4^+ + OH^-
I.....0.0786..........0........0
C......-x.............x........x
E.....0.0786-x........x........x

Substitute the E line into Kb expression for NH3 and solve for x = (OH^-), convert to pH.
Round to the correct number of significant figures.

To find the pH of NH3 (ammonia) and the resulting solution, we need to understand the reactions happening in the solution during the titration process.

A). To find the pH of NH3:
When ammonia (NH3) dissolves in water, it acts as a weak base and undergoes partial ionization according to the following equation:
NH3 + H2O ⇌ NH4+ + OH-
In this equilibrium, NH3 accepts a proton (H+) from water, forming NH4+ (ammonium ion) and releasing a hydroxide ion (OH-).

To find the pH of NH3, we need to determine the concentration of OH- ions. Since NH3 is a weak base, we can assume the concentration of NH4+ formed is approximately equal to the concentration of OH- ions.

To calculate the concentration of OH- ions, we need to determine how much NH4+ is formed during the titration. Since 20 mL of 0.100 M HCl was added to 50.0 mL of 0.150 M NH3, we need to find the amount of HCl that reacts with NH3.

By using the stoichiometry of the reaction, we can determine that the amount of HCl reacting with NH3 is equal to the amount of NH3 present. Consequently, the concentration of NH4+ formed is equal to the amount of HCl reacted because 1 mole of HCl reacts with 1 mole of NH3.

Now, let's calculate the amount of HCl reacted:
n(HCl) = C(HCl) x V(HCl)
= 0.100 M x 20 mL
= 0.002 moles

Since the stoichiometry of the reaction NH3 + HCl → NH4Cl is 1:1, the amount of NH3 reacted is also 0.002 moles.

Now, we can calculate the concentration of NH4+ formed, which is equal to the concentration of OH- ions:
C(OH-) = n(NH4+) / V(solution)
= 0.002 moles / (20 mL + 50 mL)
= 0.002 moles / 0.07 L
= 0.0286 M

Finally, using the equation for the ionization of water:
H2O ⇌ H+ + OH-
We can use the concentration of OH- ions to find the pOH of the solution, and then convert it to pH by using the equation: pH = 14 - pOH.

B). To find the pH of the resulting solution:
After the reaction between HCl and NH3, we are left with a solution containing NH4Cl (ammonium chloride) and water. Ammonium chloride is a salt formed by the neutralization of NH3 and HCl.

Since NH4Cl is a salt of a weak base (NH3) and a strong acid (HCl), the solution will be acidic. We can calculate the pH of the resulting solution by using the properties of the NH4Cl salt, such as the hydrolysis reaction:
NH4+ + H2O ⇌ NH3 + H3O+
In this reaction, NH4+ acts as an acid, donating a proton (H+) to water and forming NH3 and H3O+ (hydronium ion).

Now, we can calculate the concentration of H3O+ ions to find the pH of the solution using a similar method as in part A.

Note: It is important to use the initial volumes given to answer part A. Otherwise, the more accurate approach would be to consider the diluted volumes after mixing.