Answer:
The ratio of acid to conjugate base is outside the buffer range of 10:1.
Explanation:
The Henderson-Hasselbalch equation for a buffer is
![\text{pH} = \text{pK}_{\text{a}} + \log\dfrac{\text{[A$^{-}$]}}{\text{[HA]}}](https://tex.z-dn.net/?f=%5Ctext%7BpH%7D%20%3D%20%5Ctext%7BpK%7D_%7B%5Ctext%7Ba%7D%7D%20%2B%20%5Clog%5Cdfrac%7B%5Ctext%7B%5BA%24%5E%7B-%7D%24%5D%7D%7D%7B%5Ctext%7B%5BHA%5D%7D%7D)
A buffer should have
![\dfrac{1}{10} \leq \dfrac{\text{[A$^{-}]$}}{\text{[HA]}} \leq \dfrac{10}{1}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B10%7D%20%5Cleq%20%5Cdfrac%7B%5Ctext%7B%5BA%24%5E%7B-%7D%5D%24%7D%7D%7B%5Ctext%7B%5BHA%5D%7D%7D%20%5Cleq%20%5Cdfrac%7B10%7D%7B1%7D)
For a solution that is 1.3 mol·L⁻¹ in HF and 1.3 mmol·L⁻¹ in KF, the ratio is
The ratio of acid to conjugate base is 1000:1, which is outside the range of 10:1.
A is wrong. NF is a weak acid.
C is wrong. The two species are a conjugate acid-base pair.
D is wrong. Salts of Group 1 metals are soluble.
Answer:
a
Explanation:
because I said so and I think it's right
Since, we have the reaction as,
2Li(s) + F2(g) --> 2LiF(s)
we are only concerned with the limiting reactants. We calculate for the amount of product that can be produced with the given amount of reactants.
a. 1 g Li(1 mol / 6.941 g of Li)(2 mol LiF/2 mol Li) = 0.144 mol LiF2
1 g F2(1 mol/38 g)(2 mol LiF2/1 mol F2) = 0.052 mol LiF2
Answer: 1 g of F2
b. 10.5 g Li(1 mol/6.941 g of Li)(2 mol LiF/2 mol Li) = 1.512 mol LiF2
37.2 g F2(1 mol/38 g)(2 mol LiF2/1 mol F2) = 1.958 mol LiF2
Answer: 10.5 g of Li
c. (2.85 x 10^3 g Li)(1 mol/6.941 g of Li)(2 mol LiF/2 mol Li) = 410.60 mol LiF2
(6.79 x 10^3 g F2)(1 mol/38 g)(2 mol LiF2/1 mol F2) = 357.368 mol of LiF2
Answer: 6.79 x 10^3 g F2
