What mass of `pb^(2+)` ions is left in solution, when `50.0mL` of `0.20M Pb(NO_(3))_(2)` is added to `50.0mL` of `1.5M NAC1`?

Explanation

To find the mass of `Pb^(2+)` ions left in solution after mixing `50.0 mL` of `0.20 M Pb(NO₃)₂` with `50.0 mL` of `1.5 M NaCl`, we can follow these steps:

### Step 1: Calculate the moles of `Pb^(2+)` ions
1. **Determine the number of moles of `Pb(NO₃)₂`:**
- Molarity (M) = moles/volume (L)
- Volume of `Pb(NO₃)₂` = `50.0 mL` = `0.050 L`
- Molarity of `Pb(NO₃)₂` = `0.20 M`
- Moles of `Pb(NO₃)₂` = Molarity × Volume = `0.20 mol/L × 0.050 L = 0.010 moles`
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