Answer:
molecular so number 3. ...
Best* and are there answer choic
Answer:
The only PH range which is not covered by any of the given components of the universal indicator is 7.6-8.0
Hence the PH range 7.6-8 can't be described using universal indicator.
The 2nd ionization energy is removing a 2nd electron from that resulting cation:
<span>Li+ --> Li2+ + 1e- </span>
0.040 mol / dm³. (2 sig. fig.)
<h3>Explanation</h3>
in this question acts as a weak base. As seen in the equation in the question,
produces
rather than
when it dissolves in water. The concentration of
will likely be more useful than that of
for the calculations here.
Finding the value of
from pH:
Assume that
,
.
.
Solve for
:
![\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\text{equilibrium}} = \text{K}_b = 1.58\times 10^{-3}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D_%5Ctext%7Bequilibrium%7D%5Ccdot%5B%28%5Ctext%7BCH%7D_3%29_3%5Ctext%7BNH%7D%5E%7B%2B%7D%5D_%5Ctext%7Bequilibrium%7D%7D%7B%5B%28%5Ctext%7BCH%7D_3%29_3%5Ctext%7BN%7D%5D_%5Ctext%7Bequilibrium%7D%7D%20%3D%20%5Ctext%7BK%7D_b%20%3D%201.58%5Ctimes%2010%5E%7B-3%7D)
Note that water isn't part of this expression.
The value of Kb is quite small. The change in
is nearly negligible once it dissolves. In other words,
.
Also, for each mole of
produced, one mole of
was also produced. The solution started with a small amount of either species. As a result,
.
,
,
.