<u>Answer:</u> The pH of the buffer is 4.61
<u>Explanation:</u>
To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:
![pH=pK_a+\log(\frac{[\text{conjuagate base}]}{[\text{acid}]})](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%28%5Cfrac%7B%5B%5Ctext%7Bconjuagate%20base%7D%5D%7D%7B%5B%5Ctext%7Bacid%7D%5D%7D%29)
We are given:
= negative logarithm of acid dissociation constant of weak acid = 4.70
![[\text{conjuagate base}]}](https://tex.z-dn.net/?f=%5B%5Ctext%7Bconjuagate%20base%7D%5D%7D)
= moles of conjugate base = 3.25 moles
![[\text{acid}]](https://tex.z-dn.net/?f=%5B%5Ctext%7Bacid%7D%5D)
= Moles of acid = 4.00 moles
pH = ?
Putting values in above equation, we get:
Hence, the pH of the buffer is 4.61
Answer: 
for the reaction is -186.75 J/K
Explanation:
Change in entropy () for the given reaction under standard condition is given by-
= ![[3\times S_{rhombic}^{0}_{(s)}]+[2\times S_{H_{2}O}^{0}_{(g)}]-[2\times S_{H_{2}S}^{0}_{(g)}]-[1\times S_{SO_{2}}^{0}_{(g)}]](https://tex.z-dn.net/?f=%5B3%5Ctimes%20S_%7Brhombic%7D%5E%7B0%7D_%7B%28s%29%7D%5D%2B%5B2%5Ctimes%20S_%7BH_%7B2%7DO%7D%5E%7B0%7D_%7B%28g%29%7D%5D-%5B2%5Ctimes%20S_%7BH_%7B2%7DS%7D%5E%7B0%7D_%7B%28g%29%7D%5D-%5B1%5Ctimes%20S_%7BSO_%7B2%7D%7D%5E%7B0%7D_%7B%28g%29%7D%5D)
So = ![[3\times 31.8 J/K.mol]+[2\times 188.825 J/K.mol]-[2\times 205.79 J/K.mol]-[1\times 248.22 J/K.mol]](https://tex.z-dn.net/?f=%5B3%5Ctimes%2031.8%20J%2FK.mol%5D%2B%5B2%5Ctimes%20188.825%20J%2FK.mol%5D-%5B2%5Ctimes%20205.79%20J%2FK.mol%5D-%5B1%5Ctimes%20248.22%20J%2FK.mol%5D)
= -186.75 J/K
Answer:
1.7 × 10 ^42
Explanation:
Using Nernst equation
E°cell = RT/nF Inq
at equilibrium
Q=K
E°cell = 0.0257 /n Ink= 0.0592/n log K
Fe2+(aq)+2e−→Fe(s) E∘= −0.45 V
Ag+aq)+e−→Ag(s) E∘= 0.80 V
Fe(s)+2Ag+(aq)→Fe2+(aq)+2Ag(s)
balance the reaction
Fe → Fe²⁺ + 2e⁻ reversing for oxidation E° = 0.45 v
2 Ag⁺ +2e⁻ → 2Ag
n = 2 moles and K = equilibrium constant
E° cell = 0.80 + 0.45 = 1.25 V
E° cell = (0.0592 / n) log K
substitute the value into the equations and solve for K
(1.25 × 2) / 0.0592 = log K
42.23 = log K
k = 10^ 42.23
K = 1.7 × 10 ^42
Answer:
For a particular chemical reaction, the enthalpy of the reactants is -400 kJ. The enthalpy of the products is -390 kJ. The entropy of the reactants is 0.2 kJ/K. The entropy of the products is 0.3 kJ/K. The temperature of the reaction is 25oC. What can you conclude about this reaction?
It is exergonic
It is endergonic
it is a redox reaction
It is being catalyzed by an enzyme
