To
determine the percent ionization of the acid given, we make use of the acid
equilibrium constant (Ka) given. It is the ration of the equilibrium
concentrations of the dissociated ions and the acid. The dissociation reaction
of the HF acid would be as follows:<span>
HF = H+ + F-
The acid equilibrum constant would be expressed as follows:
Ka = [H+][F-] / [HF] = 3.5 x 10-4
To determine the equilibrium concentrations we use the ICE table,
HF
H+ F-
I 0.337 0
0
C -x +x
+x
---------------------------------------------
E 0.337-x x
x
3.5 x 10-4 = [H+][F-] / [HF]
3.5 x 10-4 = [x][x] / [0.337-x] </span>
Solving for x,
x = 0.01069 = [H+] = [F-]
percent ionization = 0.01069 / 0.337 x 100 = 3.17%
<span>Answer: 8.15s
</span><span />
<span>Explanation:
</span><span />
<span>1) A first order reaction is that whose rate is proportional to the concenration of the reactant:
</span><span />
<span>r = k [N]
</span><span />
<span>r = - d[N]/dt =
</span><span />
<span>=> -d[N]/dt = k [N]
</span><span />
<span>2) When you integrate you get:
</span><span />
<span>N - No = - kt
</span>
<span></span><span /><span>
3) Half life => N = No / 2, t = t'
</span><span />
<span>=> No - No/ 2 = kt' => No /2 = kt' => t' = (No/2) / k
</span><span />
<span>3) Plug in the data given: No = 0.884M, and k = 5.42x10⁻²M/s
</span>
<span /><span /><span>
t' = (0.884M/2) / (5.42x10⁻²M/s) = 8.15s</span>
Answer:
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Explanation:
hope this helps :P
