Answer:
There are 1.51 x 1024 molecules of carbon dioxide in 2.50 moles of carbon dioxide.
Explanation:
8.8 × 10-5 M is the [H3O+] concentration in 0.265 M HClO solution.
Explanation:
HClO is a weak acid and does not completely dissociate in water as ions.
the equation of dissociation can be written and ice table to be formed.
HClO +H2O ⇒ ClO- + H3O+
I 0.265 0 0
C -x +x +x
E 0.265-x +x +x
Now applying the equation of Ka, where Ka is given as 2.9 × 10-8.
Ka = ![\frac{[ClO-][H3O+]}{[HClO]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BClO-%5D%5BH3O%2B%5D%7D%7B%5BHClO%5D%7D)
2.9 × 10^-8 = ![\frac{[x] [x]}{[0.265-x]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Bx%5D%20%5Bx%5D%7D%7B%5B0.265-x%5D%7D)
= 7.698 x
x = 8.8 × 10-5 M
The hydronium ion concentration is 8.8 × 10-5 M in 0.265 M solution of HClO.
Answer:
a) C6H5COOH + H2O ↔ H3O+ + C6H5COO-
b) [ H3O+ ] = 2.517 E-3 M
c) pH = 2.599
Explanation:
a) balanced equation:
C6H5COOH + H2O ↔ H3O+ + C6H5COO-
⇒ Ka = ( [ H3O+ ] * [ C6H5COO- ] ) / [ C6H5COOH ] = 6.5 E-5
mass balance:
0.10 m = [ C6H5COO- ] + [ C6H5COOH ].....(1)
charge balance:
[ H3O+ ] = [ C6H5COO- ] + [ OH- ] .......[ OH- ] : comes from water, it's not significant
⇒ [ H3O+ ] = [ C6H5COO- ] .........(2)
b) (2) in (1):
⇒ 0.10 M = [ H3O+ ] + [ C6H5COOH ]
⇒ [ C6H5COOH ] = 0.10 - [ H3O+ ]
⇒ Ka = [ H3O+ ]² / ( 0.1 - [ H3O+ ] ) = 6.5 E-5
⇒ [ H3O+ ]² + 6.5 E-5 [ H3O+ ] - 6.5 E-6 = 0
⇒ [ H3O+ ] = 2.517 E-3 M
c) pH = - log [ H3O+ ]
⇒ pH = - Log ( 2.517 E-3 )
⇒ pH = 2.599
The answer you are looking for is "bromine". Hope this helps!
