A 100 mL sample of 0.10M HCl is mixed with 50mL of 0.10M NH3 (Kb=1.8x10^-5). What is the resulting pH?
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Well, if we mix HCl, which is a strong acid, with NH3, which is a weak base, it's like bringing together Batman and The Joker - a true clash of superheroes! The HCl will completely dominate the NH3 and turn it into its own acid-base slave, NH4+.
So, the reaction will be:
HCl + NH3 --> NH4+ + Cl-
Since HCl is a strong acid, it will fully dissociate in water, which means that all of the HCl turns into H+ ions. As for NH3, it will only partially dissociate because it's a weak base. But don't worry, NH3 is like an Olympic swimmer - it's pretty good at ionizing itself!
To find the pH, we need to calculate the concentration of H+ in the resulting solution. We can do this by finding the concentration of NH4+ (from NH3) and subtracting it from the initial concentration of H+ (from HCl).
Using the equation: Kb = [NH4+][OH-] / [NH3], and considering that OH- can be neglected because of its small value, we can rearrange this to: [NH4+] = Kb * [NH3].
[HCl] = 0.10 M (initial)
[NH3] = 0.10 M * (50 mL / 100 mL) = 0.05 M (final)
Kb = 1.8 x 10^-5
[NH4+] = (1.8 x 10^-5 M) * (0.05 M) = 9 x 10^-7 M
[H+] = [HCl] (initial) - [NH4+] (final) = 0.10 M - 9 x 10^-7 M
Now, to find the pH, we can plug the [H+] value into the pH formula: pH = -log[H+]
pH = -log(0.10 M - 9 x 10^-7 M)
And after doing the math, the pH of the resulting solution should be around 1.00. Just like The Joker, it's pretty acidic!
To calculate the resulting pH of the mixture, we need to determine the concentrations of the conjugate acid and base that will form when HCl and NH3 react. Then, we'll use these concentrations to calculate the pH.
First, let's calculate the concentration of the resulting conjugate acid (NH4+) and base (Cl-) after the reaction:
Given:
Volume of HCl solution (V1) = 100 mL = 0.1 L
Concentration of HCl solution (C1) = 0.10 M
Volume of NH3 solution (V2) = 50 mL = 0.05 L
Concentration of NH3 solution (C2) = 0.10 M
Since these are both strong acids and bases, they will react in a 1:1 stoichiometric ratio:
HCl(aq) + NH3(aq) → NH4+(aq) + Cl-(aq)
The moles of NH3 in the reaction can be calculated using the concentration and volume:
moles of NH3 (n2) = C2 × V2
= 0.10 M × 0.05 L
= 0.005 moles
Since the reaction is 1:1, the moles of HCl will be the same as moles of NH3:
moles of HCl (n1) = 0.005 moles
After the reaction, NH3 is consumed to form NH4+. Therefore, the remaining moles of NH3 will be:
remaining moles of NH3 = starting moles of NH3 - moles of NH3 reacted
= n2 - n1
= 0.005 - 0.005
= 0 moles
Now, let's calculate the moles of the formed conjugate acid, NH4+:
moles of NH4+ (n3) = moles of NH3 reacted
= n1
= 0.005 moles
Since we know the volume of the resulting solution, we can calculate the concentration of NH4+ and Cl-:
concentration of NH4+ = moles of NH4+ / volume of resulting solution (V3)
= n3 / (V1 + V2)
= 0.005 moles / (0.1 L + 0.05 L)
= 0.005 M / 0.15 L
= 0.0333 M
≈ 0.033 M
concentration of Cl- = moles of Cl- / volume of resulting solution (V3)
= n1 / (V1 + V2)
= 0.005 moles / 0.15 L
= 0.0333 M
≈ 0.033 M
Now, we can calculate the pOH using the Kb value of NH3:
pOH = -log10(Kb)
= -log10(1.8×10^-5)
≈ 4.74
Since water is neutral, the pH can be calculated using the equation:
pH = 14 - pOH
= 14 - 4.74
≈ 9.26
Therefore, the resulting pH of the mixture is approximately 9.26.
To find the resulting pH of the solution, we need to consider the reaction between the acid (HCl) and the base (NH3). This reaction will produce the conjugate acid of ammonia (NH4+) and the chloride ion (Cl-).
Step 1: Calculate the moles of HCl:
Given that the initial concentration of HCl is 0.10 M and the volume is 100 mL, we can calculate the number of moles of HCl using the formula:
Moles = Concentration (M) x Volume (L)
Converting mL to L:
Volume (L) = 100 mL / 1000 mL/L
Volume (L) = 0.1 L
Moles of HCl = 0.10 M x 0.1 L
Moles of HCl = 0.01 moles
Step 2: Calculate the moles of NH3:
Given that the initial concentration of NH3 is 0.10 M and the volume is 50 mL, we can calculate the number of moles of NH3:
Moles of NH3 = 0.10 M x 0.05 L
Moles of NH3 = 0.005 moles
Step 3: Determine the limiting reactant:
To find the limiting reactant, we compare the moles of HCl and NH3. The reactant with the smaller number of moles will be completely consumed, limiting the amount of product that can be formed.
In this case, NH3 has fewer moles (0.005 moles) compared to HCl (0.01 moles), so NH3 is the limiting reactant.
Step 4: Calculate the moles of NH4+ formed:
Since NH3 is the limiting reactant, it will be completely converted to NH4+. So the moles of NH4+ formed will be equal to the moles of NH3 used.
Moles of NH4+ = 0.005 moles
Step 5: Calculate the concentration of NH4+:
To find the concentration of NH4+ in the final solution, we need to consider the total volume of the solution after mixing the two solutions.
The initial volume of HCl is 100 mL, and the volume of NH3 is 50 mL. So the total volume is 100 mL + 50 mL = 150 mL.
Concentration (NH4+) = Moles of NH4+ / Total Volume (L)
Converting mL to L:
Total Volume = 150 mL / 1000 mL/L
Total Volume = 0.15 L
Concentration (NH4+) = 0.005 moles / 0.15 L
Concentration (NH4+) = 0.0333 M
Step 6: Calculate the pOH:
The pOH is calculated using the concentration of NH4+, which is a weak acid. We can consider NH4+ as the conjugate acid of NH3.
pOH = -log10 (Concentration of NH4+)
pOH = -log10 (0.0333)
pOH ≈ 1.48
Step 7: Calculate the pH:
Using the relation between pH and pOH:
pH + pOH = 14
pH + 1.48 = 14
pH = 14 - 1.48
pH ≈ 12.52
Therefore, the resulting pH of the solution is approximately 12.52.
mols HCl = M x L = ?
mols NH3 = M x L = ?
Subtract mols HCl-mols NH3,
then pH = -log(H^+)