if you start with 300g of lead acetate in 850ml of water what is the Pb^2+ concentration at a pH of 4.8?

To determine the Pb^2+ concentration at a pH of 4.8, we need to consider the equilibrium reaction that occurs between lead acetate and water. Lead acetate (Pb(CH3COO)2) is a salt that dissociates in water to release lead ions (Pb^2+) and acetate ions (CH3COO^-).

The dissociation reaction can be represented as follows:
Pb(CH3COO)2 (s) ⇌ Pb^2+ (aq) + 2CH3COO^- (aq)

Now, let's consider the pH of the solution. pH is a measure of the concentration of hydronium ions (H3O+) in a solution and is defined as the negative logarithm (base 10) of the hydronium ion concentration.

In acidic solutions, the concentration of hydronium ions is higher than that of hydroxide ions (OH^-). Therefore, we can assume that there will be no OH^- ions for the purpose of the calculation.

Given that the pH is 4.8, we can calculate the concentration of hydronium ions as follows:

[H3O+] = 10^(-pH)
[H3O+] = 10^(-4.8)

To calculate the concentration of Pb^2+ ions, we need to know the total volume of the solution. The total volume is given as 850 mL (0.85 L).

Now we can calculate the concentration of Pb^2+ ions using the following steps:

1. Calculate the moles of lead acetate (Pb(CH3COO)2):
Moles = Mass / Molar mass

The molar mass of lead acetate (Pb(CH3COO)2) is:
Pb: 207.2 g/mol
C: 12.01 g/mol
H: 1.01 g/mol
O: 16.00 g/mol

Molar mass of Pb(CH3COO)2 = (1 x 207.2) + (4 x 12.01) + (8 x 1.01) + (2 x 16.00)

2. Calculate the moles of Pb^2+ ions:
Since 1 mole of Pb(CH3COO)2 produces 1 mole of Pb^2+ ions,
Moles of Pb^2+ ions = Moles of Pb(CH3COO)2

3. Calculate the concentration of Pb^2+ ions:
Concentration (in mol/L) = Moles / Volume (in L)

Concentration of Pb^2+ ions = Moles of Pb^2+ ions / Total volume of solution

By following these steps, you can determine the concentration of Pb^2+ ions at a pH of 4.8 using the given initial mass and volume of lead acetate and water, respectively.