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:are there any abc or do you have to make an answer?
Explanation:
Answer:
22.73s
Explanation:
The reaction is a second order reaction, we know this by observing the unit of the slope.
rate constant = k = 0.056 M-1s-1
the initial concentration of BrO- [A]o = 0.80 M
time = ?
Final concentration [A]t= one-half of 0.80 M = 0.40M
1 / [A]t = kt + 1 / [A]o
1 / 0.40 = 0.056 * t + 1 / 0.80
t = (2.5 - 1.25) / 0.056
t = 22.73s
