Respuesta :
Answer: The unknown salt is NaF
Explanation:
To calculate the molarity of solution, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]
Moles of salt = 0.050 moles
Volume of solution = 0.500 L
Putting values in above equation, we get:
[tex]\text{Molarity of salt}=\frac{0.050mol}{0.500L}\\\\\text{Molarity of salt}=0.1M[/tex]
- To calculate the hydroxide ion concentration, we first calculate pOH of the solution, which is:
pH + pOH = 14
We are given:
pH = 8.08
[tex]pOH=14-8.08=5.92[/tex]
- To calculate pOH of the solution, we use the equation:
[tex]pOH=-\log[OH^-][/tex]
Putting values in above equation, we get:
[tex]5.92=-\log[OH^-][/tex]
[tex][OH^-]=10^{-5.92}=1.202\times 10^{-6}M[/tex]
The unknown salt given are formed by the combination of weak acid and strong acid which is NaOH
The chemical equation for the hydrolysis of [tex]X^-[/tex] ions follows:
[tex]X^-(aq.)+H_2O(l)\rightleftharpoons HX(aq.)+OH^-(aq.);K_b[/tex]
Initial: 0.1
At eqllm: 0.1-x x x
Concentration of [tex]OH^-=x=1.202\times 10^{-6}M[/tex]
The expression of [tex]K_b[/tex] for above equation follows:
[tex]K_b=\frac{[OH^-][HX]}{[X^-]}[/tex]
Putting values in above expression, we get:
[tex]K_b=\frac{(1.202\times 10^{-6})\times (1.202\times 10^{-6})}{(1-(1.202\times 10^{-6}))}\\\\K_b=1.445\times 10^{-11}M[/tex]
- To calculate the acid dissociation constant for the given base dissociation constant, we use the equation:
[tex]K_w=K_b\times K_a[/tex]
where,
[tex]K_w[/tex] = Ionic product of water = [tex]10^{-14}[/tex]
[tex]K_a[/tex] = Acid dissociation constant
[tex]K_b[/tex] = Base dissociation constant = [tex]1.445\times 10^{-11}[/tex]
Putting values in above equation, we get:
[tex]10^{-14}=1.445\times 10^{-11}\times K_a\\\\K_a=\frac{10^{-14}}{1.445\times 10^{-11}}=6.92\times 10^{-4}[/tex]
We know that:
[tex]K_a\text{ for HF}=6.8\times 10^{-6}[/tex]
[tex]K_a\text{ for HCl}=1.3\times 10^{6}[/tex]
[tex]K_a\text{ for HClO}=3.0\times 10^{-8}[/tex]
So, the calculated [tex]K_a[/tex] is approximately equal to the [tex]K_a[/tex] of HF
Hence, the unknown salt is NaF
Based on the calculations, the identity of this salt is sodium fluoride (NaF).
Given the following data:
- Number of moles of salt = 0.050 mol.
- Volume of solution = 0.500 Liter.
- pH of solution = 8.08.
Scientific data:
Ka for HF = [tex]6.8 \times 10^{-4}[/tex]
Ka for HCl = [tex]1.3 \times 10^{6}[/tex]
Ka for HOCl = [tex]2.8 \times 10^{-8}[/tex]
Ionic product (Kw) for water = [tex]10^{-14}[/tex]
To determine the identity of this salt:
How to calculate the molarity of solution.
First of all, we would determine the molarity of this solution.
Mathematically, the molarity of this solution is given by this formula:
[tex]Molarity = \frac{Moles\;of solute}{Volume\;of\;solution} \\\\Molarity = \frac{0.050}{0.500}[/tex]
Molarity = 0.1 M.
The hydrogen ion concentration.
Next, we would calculate the pOH and the hydrogen ion (OH) concentration as follows:
[tex]pOH = 14 -pH \\\\pOH = 14 -8.08[/tex]
pOH = 5.92.
For the OH:
[tex]pOH = -log[OH^-]\\\\5.92 = -log[OH^-]\\\\OH^-=10^{-5.92}\\\\OH = 1.202 \times 10^{-6}\;M[/tex]
For the base dissociation constant:
The hydrolysis of [tex]X^-[/tex] is given by this properly balanced chemical equation;
[tex]X^{-}(aq) + H_2O(l) \rightleftharpoons HX (aq) + OH^-(aq)[/tex]
Initial phase: 0.1
At equilibrium: 0.1 - x x x
[tex]k_b = \frac{[OH^-][HX]}{X^-} \\\\k_b =\frac{1.202 \times 10^{-6} \times 1.202 \times 10^{-6}}{1-(1.202 \times 10^{-6})} \\\\k_b = \frac{1.445 \times 10^{-12} }{0.999999 }\\\\k_b = 1.445 \times 10^{-12}[/tex]
For the acid dissociation constant:
Mathematically, acid dissociation constant is given by this formula:
[tex]k_a = \frac{k_w}{k_b} \\\\k_a = \frac{10^{-14}}{1.445 \times 10^{-12}} \\\\k_a = 6.92 \times 10^{-2}[/tex]
Therefore, the identity of this salt is sodium fluoride (NaF).
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