<u>Answer:</u>
<em>A. 10.25</em>
<em></em>
<u>Explanation:</u>
Pkb =4.77
So pka = 14 - pka = 9.23


Initial 0.50M 0 0
Change -x +x +x
Equilibrium 0.50M-x +x +x


(-x is neglected) so we get

![pH=-log[H^3 O^+]\\\\pH=-log[1.72\times10^{-5}]\\\\pH=4.76](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E3%20O%5E%2B%5D%5C%5C%5C%5CpH%3D-log%5B1.72%5Ctimes10%5E%7B-5%7D%5D%5C%5C%5C%5CpH%3D4.76)
pOH = 14 - pH
= 14 - 4.76
pOH = 9.24 is the answer
Option A - 10.25 is the answer which is close to 9.24
Answer:
Explanation:
*Since the titration is between the strong acid HCl and the strong base Ca(OH)2, the pH at the equivalent point should be 7. On interpolation, we will obtain that 9.50mL and 9.82 mL of HCl is required to completely neutralized the given Ca(OH)2 solution.
*pH at the equivalence point =7
we know that pH + pOH = 14
Hence pOH= 14-7=7
pOH= -log(OH-)
The concentration of OH-= 10-pH= 1X10-7 M
One reason for the low solubility may be the higher reaction temperature, Another reason is the common ion effect.
Answer:
D. Cyclic alkane
Explanation: there is no double or triple bond
I think the answer is choice D
Answer:
D
Explanation:
( I hope that this helps )