10.0 mL of a 0.100M Ca2+ solution was buffered at pH of 10.2. Determine the pCa values after the solution was titrated with 10.00, 20.00, and 30.00 mL of 0.0500 M EDTA solution.

Question

KCaY2- = 5.01 *10^10, alpha4 = 0.47 at pH = 10.2)

Answer 1

To determine the pCa values after titrating the solution with EDTA, we first need to understand the reaction that occurs between Ca\(^{2+}\) ions and EDTA (ethylenediaminetetraacetic acid):

\[ \text{Ca}^{2+} + \text{EDTA}^{4-} \rightleftharpoons \text{CaEDTA}^{2-} \]

In this reaction, EDTA acts as a chelating agent, forming a complex with calcium ions.

1. **Before Titration:**

Given:

- Volume of Ca\(^{2+}\) solution = 10.0 mL = 0.010 L

- Concentration of Ca\(^{2+}\) solution = 0.100 M

- pH of the solution = 10.2

First, convert pH to pCa:

\[ \text{pCa} = -\log[\text{Ca}^{2+}] \]

\[ \text{pCa} = -\log(0.100) \]

\[ \text{pCa} = -(-1) \] \[ \text{pCa} = 1 \]

So, the initial pCa value before titration is 1.

2. **After Titration with 10.00 mL of 0.0500 M EDTA:**

The balanced chemical equation for the reaction between Ca\(^{2+}\) and EDTA is 1:1, meaning 1 mole of Ca\(^{2+}\) reacts with 1 mole of EDTA. From the given information, we can calculate the moles of Ca\(^{2+}\) initially present and the moles of EDTA added.

Moles of Ca\(^{2+}\) initially = concentration \(\times\) volume = 0.100 M \(\times\) 0.010 L = 0.001 mol

Moles of EDTA added = concentration \(\times\) volume = 0.0500 M \(\times\) 0.010 L = 0.0005 mol

Since the stoichiometry is 1:1, the moles of Ca\(^{2+}\) reacted with EDTA is equal to the moles of EDTA added.

Remaining moles of Ca\(^{2+}\) = Initial moles - Moles reacted = 0.001 mol - 0.0005 mol = 0.0005 mol

Calculate the new concentration of Ca\(^{2+}\):

Concentration = Remaining moles / Total volume = 0.0005 mol / 0.010 L = 0.050 M

Now, calculate the new pCa value using the new concentration:

\[ \text{pCa} = -\log[\text{Ca}^{2+}] \]

\[ \text{pCa} = -\log(0.050) \]

\[ \text{pCa} = -(-1.30) \]

\[ \text{pCa} = 1.30 \]

So, the pCa value after titration with 10.00 mL of 0.0500 M EDTA is 1.30.

3. **After Titration with 20.00 mL of 0.0500 M EDTA:**

Following the same procedure as above:

Moles of EDTA added = 0.0500 M \(\times\) 0.020 L = 0.0010 mol

Remaining moles of Ca\(^{2+}\) = 0.001 mol - 0.0010 mol = 0 mol (all Ca\(^{2+}\) is reacted)

Concentration of Ca\(^{2+}\) = 0 mol / 0.010 L = 0 M

Since the concentration is now 0, the pCa value is not defined (pCa would be infinity).

4. **After Titration with 30.00 mL of 0.0500 M EDTA:**

Following the same procedure as above:

Moles of EDTA added = 0.0500 M \(\times\) 0.030 L = 0.0015 mol

Remaining moles of Ca\(^{2+}\) = 0.001 mol - 0.0015 mol = -0.0005 mol (negative because more EDTA than Ca\(^{2+}\) initially present)

Concentration of Ca\(^{2+}\) = -0.0005 mol / 0.010 L = -0.050 M (negative because more EDTA than Ca\(^{2+}\) initially present)

Since the concentration is negative, the pCa value is not defined (pCa would be undefined).

So, the pCa values after titration with 10.00 mL, 20.00 mL, and 30.00 mL of 0.0500 M EDTA are approximately 1.30, infinity, and undefined, respectively.