Question

Suppose 6.91 g of sodium chloride is dissolved in 100. mL of a 0.50 M aqueous solution of potassium carbonate. Calculate the final molarity of sodium cation in the solution. You can assume the volume of the solution doesn't change when the sodium chloride is dissolved in it. Round your answer to 3 significant digits. x 6 ?

Answer 1

Since the solubility of both sodium chloride and of sodium carbonate in water is very high, we can safely assume that all the added sodium cations will be in solution, regardless of the presence of potassium carbonate. Also, since NaCl yields one mole of Na⁺ per each mole of salt, the final concentration of the cation will be the same as the final concentration of the salt, which is:

$c=\frac{\frac{m}{m_{molar,NaCl}}}{V(L)}=\frac{\frac{6.91g}{58.44g/mol}}{0.100L}=1.18\frac{mol}{L}=1.18M$