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.
Answer:
The closest shell to the nucleus is called the "1 shell" (also called the "K shell"), followed by the "2 shell" (or "L shell"), then the "3 shell" (or "M shell"), and so on farther and farther from the nucleus.
Answer:
potassium hydrogen phthalate KHP MOLAR MASS = 204.233 glmol
to get 1000 ml
Molar concentration = Mass concentration/Molar Mass
mass concentration = molar concentration x molar mass
mass concentration=0.1 M,
molar mass= 204.233 g/mol
so to get 1L
mass conc = 204.233 x 0.1
= 20.4233g for 1L or 1000 ml
to get 6.00 ml
if 20.4233g is for 1000ml
then to 6.00 ml
= 20.4233 x 6 / 1000
= 0.123g for 6.00 ml
according to the equation below
NaOH(aq) + KHC8H4O4(aq) --> KNaC8H4O4(aq) + H2O(l)
number of moles of NaOH is equal to that of KHP
so the same amount will be needed too, which is
= 0.123g
The volume that will occupy at STP is calculated as follows
by use of ideal gas equation
that is PV=nRT where n is number of moles calculate number of moles
n= PV/RT
p=0.75 atm
V=6.0 L
R = 0.0821 L.atm/k.mol
T= 35 +273= 308k
n=?
n= (o.75 atm x 6.0 L)/( 0.0821 L.atm/k.mol x 308 k)= 0.178 moles
Agt STP 1 mole= 22.4 L what obout 0.178 moles
= 22.4 x0.178moles/ 1moles =3.98 L( answer C)
