<u>Answer:</u> The unknown salt is NaF
<u>Explanation:</u>
To calculate the molarity of solution, we use the equation:
![\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}](https://tex.z-dn.net/?f=%5Ctext%7BMolarity%20of%20the%20solution%7D%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20solute%7D%7D%7B%5Ctext%7BVolume%20of%20solution%20%28in%20L%29%7D%7D)
Moles of salt = 0.050 moles
Volume of solution = 0.500 L
Putting values in above equation, we get:
![\text{Molarity of salt}=\frac{0.050mol}{0.500L}\\\\\text{Molarity of salt}=0.1M](https://tex.z-dn.net/?f=%5Ctext%7BMolarity%20of%20salt%7D%3D%5Cfrac%7B0.050mol%7D%7B0.500L%7D%5C%5C%5C%5C%5Ctext%7BMolarity%20of%20salt%7D%3D0.1M)
- To calculate the hydroxide ion concentration, we first calculate pOH of the solution, which is:
pH + pOH = 14
We are given:
pH = 8.08
![pOH=14-8.08=5.92](https://tex.z-dn.net/?f=pOH%3D14-8.08%3D5.92)
- To calculate pOH of the solution, we use the equation:
![pOH=-\log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5BOH%5E-%5D)
Putting values in above equation, we get:
![5.92=-\log[OH^-]](https://tex.z-dn.net/?f=5.92%3D-%5Clog%5BOH%5E-%5D)
![[OH^-]=10^{-5.92}=1.202\times 10^{-6}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D10%5E%7B-5.92%7D%3D1.202%5Ctimes%2010%5E%7B-6%7DM)
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
ions follows:
![X^-(aq.)+H_2O(l)\rightleftharpoons HX(aq.)+OH^-(aq.);K_b](https://tex.z-dn.net/?f=X%5E-%28aq.%29%2BH_2O%28l%29%5Crightleftharpoons%20HX%28aq.%29%2BOH%5E-%28aq.%29%3BK_b)
<u>Initial:</u> 0.1
<u>At eqllm:</u> 0.1-x x x
Concentration of ![OH^-=x=1.202\times 10^{-6}M](https://tex.z-dn.net/?f=OH%5E-%3Dx%3D1.202%5Ctimes%2010%5E%7B-6%7DM)
The expression of
for above equation follows:
![K_b=\frac{[OH^-][HX]}{[X^-]}](https://tex.z-dn.net/?f=K_b%3D%5Cfrac%7B%5BOH%5E-%5D%5BHX%5D%7D%7B%5BX%5E-%5D%7D)
Putting values in above expression, we get:
![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](https://tex.z-dn.net/?f=K_b%3D%5Cfrac%7B%281.202%5Ctimes%2010%5E%7B-6%7D%29%5Ctimes%20%281.202%5Ctimes%2010%5E%7B-6%7D%29%7D%7B%281-%281.202%5Ctimes%2010%5E%7B-6%7D%29%29%7D%5C%5C%5C%5CK_b%3D1.445%5Ctimes%2010%5E%7B-11%7DM)
- To calculate the acid dissociation constant for the given base dissociation constant, we use the equation:
![K_w=K_b\times K_a](https://tex.z-dn.net/?f=K_w%3DK_b%5Ctimes%20K_a)
where,
= Ionic product of water = ![10^{-14}](https://tex.z-dn.net/?f=10%5E%7B-14%7D)
= Acid dissociation constant
= Base dissociation constant = ![1.445\times 10^{-11}](https://tex.z-dn.net/?f=1.445%5Ctimes%2010%5E%7B-11%7D)
Putting values in above equation, we get:
![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}](https://tex.z-dn.net/?f=10%5E%7B-14%7D%3D1.445%5Ctimes%2010%5E%7B-11%7D%5Ctimes%20K_a%5C%5C%5C%5CK_a%3D%5Cfrac%7B10%5E%7B-14%7D%7D%7B1.445%5Ctimes%2010%5E%7B-11%7D%7D%3D6.92%5Ctimes%2010%5E%7B-4%7D)
We know that:
![K_a\text{ for HF}=6.8\times 10^{-6}](https://tex.z-dn.net/?f=K_a%5Ctext%7B%20for%20HF%7D%3D6.8%5Ctimes%2010%5E%7B-6%7D)
![K_a\text{ for HCl}=1.3\times 10^{6}](https://tex.z-dn.net/?f=K_a%5Ctext%7B%20for%20HCl%7D%3D1.3%5Ctimes%2010%5E%7B6%7D)
![K_a\text{ for HClO}=3.0\times 10^{-8}](https://tex.z-dn.net/?f=K_a%5Ctext%7B%20for%20HClO%7D%3D3.0%5Ctimes%2010%5E%7B-8%7D)
So, the calculated
is approximately equal to the
of HF
Hence, the unknown salt is NaF