A kilogram is a unit of mass. A meter is a unit of measurement, a Liter is volume and a kelvin is temperature
Answer:
![[H^+]=0.000285](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.000285)
![pH=3.55](https://tex.z-dn.net/?f=pH%3D3.55)
Explanation:
In this, we can with the <u>ionization equation</u> for the hydrazoic acid (
). So:
![HN_3~~H^+~+~N_3^-](https://tex.z-dn.net/?f=HN_3~%3C-%3E~H%5E%2B~%2B~N_3%5E-)
Now, due to the Ka constant value, we have to use the whole equilibrium because this <u>is not a strong acid</u>. So, we have to write the <u>Ka expression</u>:
![Ka=\frac{[H^+][N_3^-]}{[HN_3]}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B%5BH%5E%2B%5D%5BN_3%5E-%5D%7D%7B%5BHN_3%5D%7D)
For each mol of
produced we will have 1 mol of
. So, we can use <u>"X" for the unknown</u> values and replace in the Ka equation:
![Ka=\frac{X*X}{[HN_3]}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7BX%2AX%7D%7B%5BHN_3%5D%7D)
Additionally, we have to keep in mind that
is a reagent, this means that we will be <u>consumed</u>. We dont know how much acid would be consumed but we can express a<u> subtraction from the initial value</u>, so:
![Ka=\frac{X*X}{0.004-X}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7BX%2AX%7D%7B0.004-X%7D)
Finally, we can put the ka value and <u>solve for "X"</u>:
![2.2X10^-^5=\frac{X*X}{0.004-X}](https://tex.z-dn.net/?f=2.2X10%5E-%5E5%3D%5Cfrac%7BX%2AX%7D%7B0.004-X%7D)
![2.2X10^-^5=\frac{X^2}{0.004-X}](https://tex.z-dn.net/?f=2.2X10%5E-%5E5%3D%5Cfrac%7BX%5E2%7D%7B0.004-X%7D)
![X= 0.000285](https://tex.z-dn.net/?f=X%3D%200.000285)
So, we have a concentration of 0.000285 for
. With this in mind, we can calculate the <u>pH value</u>:
![pH=-Log[H^+]=-Log[0.000285]=3.55](https://tex.z-dn.net/?f=pH%3D-Log%5BH%5E%2B%5D%3D-Log%5B0.000285%5D%3D3.55)
I hope it helps!
Not 100% sure, but I think it’s Water, hope this helps
False , it will make it lighter