Question

Calculate the sodium ion concentration when 70.0 mL of 3.0 M sodium carbonate is added to 30.0 mL of 1.0 M sodium bicarbonate.

Short answer

Concentration of sodium ion is 4.5 M

Step-by-step solution

Given Data

Given that , volume of sodium carbonate is 70mL and volume of sodium

bicarbonate is 30mL .

Number of moles of sodium in 1 mol of sodium carbonate is 2 mol

Number of moles of sodium in 1 mol of sodium bicarbonate is 1 mol

Moles of sodium bicarbonate = molarity $\times$volume of the solution in L

=$1\times \frac{30}{1000}=0.03\text{mol}$

 Calculate the sodium ion concentration

Moles of sodium in 0.03 moles of sodium bicarbonate =$0.03\times 1=0.03\text{mol}$

Similarly, moles of sodium carbonate =$3\times \frac{70}{1000}=0.21\text{mol}$

Moles of sodium in 0.21 moles of sodium carbonate =$0.21\times 2=0.42\text{mol}$

Total number of moles of sodium = 0.03 + 0.42 = 0.45 mol

Volume of solution of sodium carbonate and sodium bicarbonate in liters

=$\frac{(30+70)}{1000}=0.1L$

Concentration of sodium in the solution=$\frac{0.45}{0.1}$ = 4.5 M