Change in their environment
Answer:
Explanation:
Hello,
In this case, for a concentration of 0.42 M of benzoic acid whose Ka is 6.3x10⁻⁵ in 0.33 M sodium benzoate, we use the Henderson-Hasselbach equation to compute the required pH:
![pH=pKa+log(\frac{[base]}{[acid]} )](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%20%29)
Whereas the concentration of the base is 0.33 M and the concentration of the acid is 0.42 M, thereby, we obtain:
![pH=-log(Ka)+log(\frac{[base]}{[acid]} )\\\\pH=-log(6.3x10^{-5})+log(\frac{0.33M}{0.42M} )\\\\pH=4.1](https://tex.z-dn.net/?f=pH%3D-log%28Ka%29%2Blog%28%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%20%29%5C%5C%5C%5CpH%3D-log%286.3x10%5E%7B-5%7D%29%2Blog%28%5Cfrac%7B0.33M%7D%7B0.42M%7D%20%29%5C%5C%5C%5CpH%3D4.1)
Regards.
Answer:
<h2>2.45 g/mL</h2>
Explanation:
The density of a substance can be found by using the formula
We have
We have the final answer as
<h3>2.45 g/mL</h3>
Hope this helps you
Answer: With the cold front, warm air is rapidly forced upward (like the shavings) in advance of the actual front (the “cutter”), creating towering cumulus clouds, some hard showers and quite possibly a few gusty thunderstorms followed by a push of cooler and drier air in its wake.
Explanation:
Answer:
Ka = 0.1815
Explanation:
Chromic acid
pH = ?
Concentration = 0.078 M
Ka = ?
HCl
conc. = 0.059M
pH = -log(H+)
pH = -log(0.059) = 1.23
pH of chromic acid = 1.23
Step 1 - Set up Initial, Change, Equilibrium table;
H2CrO4 ⇄ H+ + HCrO4−
Initial - 0.078M 0 0
Change : -x +x +x
Equilibrium : 0.078-x x x
Step 2- Write Ka as Ratio of Conjugate Base to Acid
The dissociation constant Ka is [H+] [HCrO4−] / [H2CrO4].
Step 3 - Plug in Values from the Table
Ka = x * x / 0.078-x
Step 4 - Note that x is Related to pH and Calculate Ka
[H+] = 10^-pH.
Since x = [H+] and you know the pH of the solution,
you can write x = 10^-1.23.
It is now possible to find a numerical value for Ka.
Ka = (10^-1.23))^2 / (0.078 - 10^-1.23) = 0.00347 / 0.0191156
Ka = 0.1815
