theoretical pH
14-0.5(pKb-log[base]) = ?
please help
What's the question? Something is missing.
sorry 0.1 M NH4OH has pH 9.8 so I need the theoretical pH for 14 - 0.5(pKb-log [base])=?
also 0.1 M HAc pH=2.5 theoretical pH
0.5 (pKa-log [acid])= ?
Gratefull for help
To solve the given equation, let's go step by step:
1. First, we need to understand the components of the equation:
- pKb: It represents the negative logarithm (base 10) of the base dissociation constant (Kb).
- [base]: It represents the concentration of the base being considered.
2. Now, let's rewrite the equation by separating it into different parts:
- pKb-log[base] = (14 - 0.5(pKb-log[base]))
3. Next, let's simplify the equation by making the pKb and log[base] terms on the right-hand side:
- pKb-log[base] + log[base] = 14 - 0.5pKb
4. Since the log[base] term and -log[base] term cancel each other, we get:
- pKb = 14 - 0.5pKb
5. Moving the pKb term to the left-hand side and the constant terms to the right-hand side, we have:
- 1.5pKb = 14
6. Finally, we can solve for pKb by dividing both sides of the equation by 1.5:
- pKb = 14 / 1.5
By following these steps, we have determined that pKb is equal to 9.33.