<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:yes
Explanation:
this is because the distance doesnt matter
Answer:
yes
Explanation:
molecule, a group of two or more atoms that form the smallest identifiable unit into which a pure substance
Answer:
V = 134.5 L
Explanation:
Given data:
Number of moles of KClO₃ = 4 mol
Litters of oxygen produced at STP = ?
Solution:
Chemical equation:
2KClO₃ → 2KCl + 3O₂
Now we will compare the moles of KClO₃ with oxygen.
KClO₃ : O₂
2 : 3
4 ; 3/2×4 = 6 mol
Litters of oxygen at STP:
PV = nRT
V = nRT/P
V = 6 mol × 0.0821 atm.L/mol.K × 273 K / 1atm
V = 134.5 L / 1
V = 134.5 L
