Answered: Calculate the molarity of sodium ion in…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Question

Calculate the molarity of sodium ion in a solution made by mixing 4.63 mL of 0.398 M sodium chloride with 550.00 mL of 6.55 × 10⁻² M sodium sulfate. Assume volumes are additive.

Expert Solution

Step 1

The given volume of NaCl solution = 4.63 ml = 0.00463 L

The given molarity of NaCl solution = 0.398 M = 0.398 mol/L

$\begin{array}{rcl} Number of moles of NaCl & = & (molarity) \times (volume) \\ = & (0.398 mol / L) \times (0.00463 L) \\ = & 0.0018 mol \end{array}$

$NaCl (aq) \overset{}{\to} {Na}^{+} (aq) + {Cl}^{-} (aq)$

$\begin{array}{rcl} 1 mol of NaCl contributes to the solution & = & 1 mol of {Na}^{+} ions \\ So, (0.0018 mol) of NaCl would contribute to the solution & = & (0.0018 mol) of {Na}^{+} ions . . . . . . . . . . . (1) \end{array}$