Pk=3 pH=4,7 C-?

Question 1

Задача Задача №2400636

А

Аноним

03 декабря 2024 06:22 · публичный вопрос

А

Аноним 03 декабря 2024 06:22

Pk=3 pH=4,7 C-?

Answer

To find the concentration of an acid (C) from the given \( pK_a \) and \( pH \), you can follow these steps.

The \( pK_a \) is given as 3, which means that: \[ K_a = 10^{-pK_a} = 10^{-3} = 0.001 \]

The \( pH \) is given as 4.7, which allows us to find the concentration of hydrogen ions \([H^+]\): \[ [H^+] = 10^{-pH} = 10^{-4.7} \approx 2.0 \times 10^{-5} \, \text{M} \]

Assuming you have a weak acid that dissociates according to the equation: \[ HA \rightleftharpoons H^+ + A^- \]

We can set up the expression for \(K_a\): \[ K_a = \frac{[H^+][A^-]}{[HA]} \]

Where:

\([H^+]\) is the concentration of hydrogen ions.
\([A^-]\) is the concentration of the conjugate base.
\([HA]\) is the concentration of the undissociated acid.

In a solution at equilibrium, since we assume the acid is weak and only partially dissociates, we can equate \([H^+] = [A^-]\) at equilibrium.

Let the initial concentration of the weak acid be \(C\). The change in concentration when it dissociates is \(x\) (where \(x = [H^+] = 2.0 \times 10^{-5}\)). Thus, at equilibrium, we have: \[ [HA] \approx C - x \quad \text{(since } x \text{ is small compared to } C\text{)} \]

Substituting these into the expression for \(K_a\): \[ K_a = \frac{(2.0 \times 10^{-5})(2.0 \times 10^{-5})}{C - 2.0 \times 10^{-5}} \]

\[ 0.001 = \frac{(2.0 \times 10^{-5})^2}{C - 2.0 \times 10^{-5}} \]

Solving for \(C\): \[ C - 2.0 \times 10^{-5} = \frac{(2.0 \times 10^{-5})^2}{0.001} \]

Calculate \((2.0 \times 10^{-5})^2 = 4.0 \times 10^{-10}\):

\[ C - 2.0 \times 10^{-5} = \frac{4.0 \times 10^{-10}}{0.001} = 4.0 \times 10^{-7} \]

Thus: \[ C = 4.0 \times 10^{-7} + 2.0 \times 10^{-5} \]

Calculating that gives: \[ C \approx 2.04 \times 10^{-5} \, \text{M} \]

So, the concentration of the acid \( C \) is approximately \( 2.04 \times 10^{-5} \, \text{M} \).

Question 2

Pk=3 pH=4,7 C-?

Answer

To find the concentration of an acid (C) from the given \( pK_a \) and \( pH \), you can follow these steps. The \( pK_a \) is given as 3, which means that: \[ K_a = 10^{-pK_a} = 10^{-3} = 0.001 \] The \( pH \) is given as 4.7, which allows us to find the concentration of hydrogen ions \([H^+]\): \[ [H^+] = 10^{-pH} = 10^{-4.7} \approx 2.0 \times 10^{-5} \, \text{M} \] Assuming you have a weak acid that dissociates according to the equation: \[ HA \rightleftharpoons H^+ + A^- \] We can set up the expression for \(K_a\): \[ K_a = \frac{[H^+][A^-]}{[HA]} \] Where: - \([H^+]\) is the concentration of hydrogen ions. - \([A^-]\) is the concentration of the conjugate base. - \([HA]\) is the concentration of the undissociated acid. In a solution at equilibrium, since we assume the acid is weak and only partially dissociates, we can equate \([H^+] = [A^-]\) at equilibrium. Let the initial concentration of the weak acid be \(C\). The change in concentration when it dissociates is \(x\) (where \(x = [H^+] = 2.0 \times 10^{-5}\)). Thus, at equilibrium, we have: \[ [HA] \approx C - x \quad \text{(since } x \text{ is small compared to } C\text{)} \] Substituting these into the expression for \(K_a\): \[ K_a = \frac{(2.0 \times 10^{-5})(2.0 \times 10^{-5})}{C - 2.0 \times 10^{-5}} \] \[ 0.001 = \frac{(2.0 \times 10^{-5})^2}{C - 2.0 \times 10^{-5}} \] Solving for \(C\): \[ C - 2.0 \times 10^{-5} = \frac{(2.0 \times 10^{-5})^2}{0.001} \] Calculate \((2.0 \times 10^{-5})^2 = 4.0 \times 10^{-10}\): \[ C - 2.0 \times 10^{-5} = \frac{4.0 \times 10^{-10}}{0.001} = 4.0 \times 10^{-7} \] Thus: \[ C = 4.0 \times 10^{-7} + 2.0 \times 10^{-5} \] Calculating that gives: \[ C \approx 2.04 \times 10^{-5} \, \text{M} \] So, the concentration of the acid \( C \) is approximately \( 2.04 \times 10^{-5} \, \text{M} \).