Answer:
(d) 4.5 M
Explanation:
We have 70.0 mL of 3.0-molar Na₂CO₃. The moles of Na₂CO₃ are:
70.0 × 10⁻³ L × 3.0 mol/L = 0.21 mol
There are 2 moles of Na⁺ in 1 mole of Na₂CO₃. The moles of Na⁺ in 0.21 moles of Na₂CO₃ are 2 × 0.21 mol = 0.42 mol
We have 30.0 mL of 1.0-molar NaHCO₃. The moles of NaHCO₃ are:
30.0 × 10⁻³ L × 1.0 mol/L = 0.030 mol
There is 1 mole of Na⁺ in 1 mole of NaHCO₃. The moes of Na⁺ in 0.030 moles of NaHCO₃ are 1 × 0.030 mol = 0.030 mol
The total number of moles of Na⁺ is 0.42 mol + 0.030 mol = 0.45 mol
The total volume is 70.0 mL + 30.0 mL = 100.0 mL
The concentration of Na⁺ is
0.45 mol / 0.1000 L = 4.5 M