Answer:
Explanation:
Hello,
In this case, since the pH defines the concentration of hydrogen:
![pH=-log([H^+])](https://tex.z-dn.net/?f=pH%3D-log%28%5BH%5E%2B%5D%29)
![[H^+]=10^{-pH}=10^{-3.4}=3.98x10^{-4}](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B-pH%7D%3D10%5E%7B-3.4%7D%3D3.98x10%5E%7B-4%7D)
And the percent ionization is:
![\% \ ionization=\frac{[H^+]}{[HA]}*100\%](https://tex.z-dn.net/?f=%5C%25%20%5C%20ionization%3D%5Cfrac%7B%5BH%5E%2B%5D%7D%7B%5BHA%5D%7D%2A100%5C%25)
We compute the concentration of the acid, HA:
![[HA]=\frac{[H^+]}{\% \ ionization}*100\%=\frac{3.98x10^{-4}}{66\%} *100\%\\\\](https://tex.z-dn.net/?f=%5BHA%5D%3D%5Cfrac%7B%5BH%5E%2B%5D%7D%7B%5C%25%20%5C%20ionization%7D%2A100%5C%25%3D%5Cfrac%7B3.98x10%5E%7B-4%7D%7D%7B66%5C%25%7D%20%20%2A100%5C%25%5C%5C%5C%5C)
![[HA]=6.03x10^{-4}](https://tex.z-dn.net/?f=%5BHA%5D%3D6.03x10%5E%7B-4%7D)
Thus, the Ka is:
![Ka=\frac{[H^+][A^-]}{[HA]}=\frac{3.98x10^{-4}*3.98x10^{-4}}{6.03x10^{-4}}\\ \\Ka=2.63x10^{-4}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B%5BH%5E%2B%5D%5BA%5E-%5D%7D%7B%5BHA%5D%7D%3D%5Cfrac%7B3.98x10%5E%7B-4%7D%2A3.98x10%5E%7B-4%7D%7D%7B6.03x10%5E%7B-4%7D%7D%5C%5C%20%20%5C%5CKa%3D2.63x10%5E%7B-4%7D)
So the pKa is:
Regards.
Yes what do u need help with
____NaNO3 + ___PbO --> ___Pb(NO3)2 + ___Na[2]O
To balace the eqaution, you need to have the same number of atoms for each element on both the reactant (left) and product (right) side.
To start off, you wanna know the number of atoms in each element on both sides, so take it apart:
[reactants] [product]
Na- 1 Na- 2
N- 1 N- 2(it's 2 because the the subscript [2] is outside of the parenthesis)
O- 4 O- 7 (same reason as above)
Pb- 1 Pb- 1
Na is not balanced out, so add a coefficient to make it the same on both sides.In this case, multiply by 2:
2NaNO3
Now Na is balanced, but the N and O are also effected by this, so they also have to be multiplied by 2 and they become:
Na- 2 Na- 2
N- 2 N- 2 (it balanced out)
O- 7 (coefficient times subscript, plus lone O) O- 7 (balanced out)
Pb was already balanced so no need to mess with it, just put a 1 where needed (it doesn't change anything).
Now to put it back together, it will look like this:
2NaNO3 + 1PbO --> 1Pb(NO3)2 + 1Na[2]O
