You must know and use the formula for pH.
pH = - log [H3O+], where [H3O+] is the molar concentration of hydronium ion.
So, when pH is 8.0 => 8.0 = - log [H3O+] and you can use antilogarithm (the inverse function of logarithm) to find [H3O+], in this way:
[H3O+] = 10^-8 = 1 * 10 ^-8 M
When, pH = 7.0 =>
7.0 = - log [H3O+] => [H3O+] = 1 * 10^ -7 M
Answer: 1*10^-7 mole / liter
Answer:
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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)
<em>pH = 5.83</em>
