The reaction between the reactants would be:

CH₃NH₂ + HCl ↔ CH₃NH₃⁺ + Cl⁻

Let the conjugate acid undergo hydrolysis. Then, apply the ICE approach.

CH₃NH₃⁺ + H₂O → H₃O⁺ + CH₃NH₂
I 0.11 0 0
C -x +x +x
E 0.11 - x x x

Ka = [H₃O⁺][CH₃NH₂]/[CH₃NH₃⁺]

Since the given information is Kb, let's find Ka in terms of Kb.

Ka = Kw/Kb, where Kw = 10⁻¹⁴

So,
Ka = 10⁻¹⁴/5×10⁻⁴ = 2×10⁻¹¹ = [H₃O⁺][CH₃NH₂]/[CH₃NH₃⁺]
2×10⁻¹¹ = [x][x]/[0.11-x]
Solving for x,
x = 1.483×10⁻⁶ = [H₃O⁺]

Since pH = -log[H₃O⁺],
pH = -log(1.483×10⁻⁶)
pH = 5.83

4 0

3 years ago

A solution is prepared by adding 0.0231moles of H3O+ ions to 3.33L of water. What is the pH of this solution

MrMuchimi

Answer:

2.15

Explanation:

For this question, we have to remember the pH formula:

$pH~=~-Log[H_3O^+]$

By definition, the pH value is calculated when we do the -Log of the concentration of the hydronium ions ( $H_3O^+$ ). So, the next step is the calculation of the concentration of the hydronium ions. For this, we have to use the molarity formula:

$M=\frac{mol}{L}$

We already know the number of moles (0.0231 moles) and the volume (3.33 L). So, we can plug the values into the molarity formula:

$M=\frac{0.0231~moles}{3.33~L}=0.00693~M$

With this value, now we can calculate the pH value:

$pH~=~-Log[0.00693~M]~=~2.15$

The pH would be 2.15

I hope it helps!

8 0

3 years ago