Lol I don’t have snap. But I’m only texting this cause this app wants me to answer some questions. But! Lol have a great day/night
<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
E = mct
Energy = (mass) x (specific heat capacity of water) x (change in temp)
585.24 = 53.2 x 4.2 x (X-24.15)
585.24 divided by 53.2 divided by 4.2 = X - 24.15
2.62 = X - 24.15
X= 26.77degrees C
(Specific heat capacity for water is 4.2 but is different for other liquids)