A soluction of 100 mL of 0.2 M NH3 is titratedwith 50 mL of 0.5 M HNO3. What is the
pH of the resulting solution? Kb for NH3 is
1.8 × 10−5
1. 5.30
2. 1.00
3. 1.48
4. 10.3
NH3 + HNO3 ==> NH4NO3 + H2O
millimols NH3 = 100 x 0.2 = 20
mmols HNO3 = 50 x 0.5 = 25
So you've added enough HNO3 to neutralize all of the NH3 and have an excess of 5 mmols HNO3 left. That's in 100 + 50 = 150 mL
M HNO3 = mmols/ml = ?M
To find the pH of the resulting solution, we need to consider the reaction between NH3 (ammonia) and HNO3 (nitric acid).
The balanced equation for the reaction is:
NH3 + HNO3 -> NH4+ + NO3-
In this reaction, NH3 acts as a base, accepting a proton (H+) from HNO3, which acts as an acid. The resulting products are NH4+ (ammonium ion) and NO3- (nitrate ion).
To find the pH of the resulting solution, we need to calculate the concentration of the ammonium ion (NH4+). This can be done using the stoichiometry of the reaction and the initial concentrations of NH3 and HNO3.
First, let's calculate the moles of NH3 and HNO3 used in the reaction:
moles of NH3 = volume of NH3 solution (in L) * concentration of NH3
= 0.1 L * 0.2 M
= 0.02 moles of NH3
moles of HNO3 = volume of HNO3 solution (in L) * concentration of HNO3
= 0.05 L * 0.5 M
= 0.025 moles of HNO3
Since the stoichiometric ratio between NH3 and NH4+ is 1:1, the moles of NH4+ formed will be equal to the moles of NH3 used.
moles of NH4+ = moles of NH3 used
= 0.02 moles
Now, let's calculate the concentration of NH4+ in the final solution:
concentration of NH4+ = moles of NH4+ / total volume of final solution (in L)
= 0.02 moles / (0.1 L + 0.05 L)
= 0.02 moles / 0.15 L
= 0.133 M
Next, we need to find the pOH of the solution using the Kb value for NH3:
pOH = -log10(Kb) = -log10(1.8 x 10^-5)
Using this value of pOH, we can find the pH of the solution:
pH = 14 - pOH
Now, let's calculate pOH and pH:
pOH = -log10(1.8 x 10^-5) ≈ 4.74
pH = 14 - pOH = 14 - 4.74 ≈ 9.26
So, the pH of the resulting solution is approximately 9.26.
None of the given options (1. 5.30, 2. 1.00, 3. 1.48, 4. 10.3) are the correct answer.
To find the pH of the resulting solution, you need to calculate the concentration of the resulting solution after the reaction between NH3 (ammonia) and HNO3 (nitric acid) takes place.
Let's start by determining the number of moles of NH3 and HNO3 used in the reaction.
Moles of NH3 = volume of solution (in liters) x molarity of NH3
Moles of NH3 = (100 mL / 1000 mL/L) x 0.2 M
Moles of NH3 = 0.02 moles
Moles of HNO3 = volume of solution (in liters) x molarity of HNO3
Moles of HNO3 = (50 mL / 1000 mL/L) x 0.5 M
Moles of HNO3 = 0.025 moles
Since HNO3 is a strong acid, it will react completely to form H+ ions and NO3- ions. Therefore, after the reaction, the number of moles of H+ ions in the solution will be equal to the moles of HNO3 used.
The balanced chemical equation for the reaction between NH3 and HNO3 is:
NH3 + HNO3 --> NH4+ + NO3-
From the balanced equation, we can see that one mole of NH3 reacts with one mole of HNO3 to form one mole of NH4+ ions.
Therefore, the number of moles of NH4+ ions formed after the reaction is 0.025 moles.
Now, let's calculate the concentration of NH4+ ions in the resulting solution.
Concentration of NH4+ ions = moles of NH4+ / volume of resulting solution (in liters)
Concentration of NH4+ ions = 0.025 moles / (100 mL + 50 mL) / 1000 mL/L
Concentration of NH4+ ions = 0.025 moles / 0.15 L
Concentration of NH4+ ions = 0.1667 M
Since NH4+ is the conjugate acid of NH3, we can use the given Kb value for NH3 to find the pOH of the resulting solution. Then, we can convert pOH to pH using the equation pH = 14 - pOH.
pOH = -log(Kb) + log(concentration of NH4+) (using the equation Kb = [NH4+][OH-] / [NH3])
pOH = -log(1.8 × 10−5) + log(0.1667)
pOH = 4.74 - 0.778
pOH = 3.962
pH = 14 - pOH
pH = 14 - 3.962
pH ≈ 10.038
Therefore, the pH of the resulting solution is approximately 10.038.