Answer:
6.68 X 10^-11
Explanation:
From the second Ka, you can calculate pKa = -log (Ka2) = 6.187
The pH at the second equivalence point (8.181) will be the average of pKa2 and pKa3. So,
8.181 = (6.187 + pKa3) / 2
Solving gives pKa3 = 10.175, and Ka3 = 10^-pKa3 = 6.68 X 10^-11
The last row going across
I don’t want to be wrong but I’m gonna say 2p
To fully understand the problem, we use the ICE table to identify the concentration of the species. We calculate as follows:
Ka = 2.0 x 10^-9 = [H+][OBr-] / [HOBr]
HOBr = 0.50 M
KOBr = 0.30 M = OBr-
<span> HOBr + H2O <-> H+ + OBr- </span>
<span>I 0.50 - 0 0.30 </span>
<span>C -x x x
</span>---------------------------------------------
<span>E(0.50-x) x (0.30+x) </span>
<span>Assuming that the value of x is small as compared to 0.30 and 0.50 </span>
<span>Ka = 2.0 x 10^-9 = x (0.30) / 0.50) </span>
<span>x = 3.33 x 10^-9 = H+</span>
pH = 8.48
Answer:
a. -0.63 V
b. No
Explanation:
Step 1: Given data
- Standard reduction potential of the anode (E°red): -1.33 V
- Minimum standard cell potential (E°cell): 0.70 V
Step 2: Calculate the required standard reduction potential of the cathode
The galvanic cell must provide at least 0.70V of electrical power, that is:
E°cell > 0.70 V [1]
We can calculate the standard reduction potential of the cathode (E°cat) using the following expression.
E°cell = E°cat - E°an [2]
If we combine [1] and [2], we get,
E°cat - E°an > 0.70 V
E°cat > 0.70 V + E°an
E°cat > 0.70 V + (-1.33 V)
E°cat > -0.63 V
The minimum E°cat is -0.63 V and there is no maximum E°cat.
