How many moles of Na+ are present in 343 mL of a 1.27 M solution of Na2SO4 quizlet?

To calculate the moles of Na+ in a sodium sulfate (Na2SO4) solution, we need to use the molarity (M) of the solution along with the volume in liters. The molarity tells us the number of moles of solute (in this case Na2SO4) per liter of solution. Let’s break this question down step-by-step:

1. Write the balanced equation for sodium sulfate

Sodium sulfate dissociates into 2 sodium ions (Na+) and 1 sulfate ion (SO42-) when dissolved in water:

Na2SO4(s) → 2Na+(aq) + SO42-(aq)

2. Identify the given values

  • Molarity (M) of Na2SO4 solution = 1.27 M
  • Volume (V) of Na2SO4 solution = 343 mL

Note that we need to convert the volume into liters because molarity is defined per liter.

3. Convert volume into liters

343 mL x (1 L / 1000 mL) = 0.343 L

4. Use the molarity equation to calculate moles of Na2SO4

Molarity (M) = moles solute / liters solution

Moles Na2SO4 = M x L

= 1.27 mol/L x 0.343 L
= 0.436 mol Na2SO4

5. Use the mole ratio from the balanced equation to relate moles of Na2SO4 to moles Na+

From the balanced equation:

2 mol Na+ / 1 mol Na2SO4

So if we have 0.436 mol Na2SO4, then we must have:

0.436 mol Na2SO4 x (2 mol Na+ / 1 mol Na2SO4) = 0.872 mol Na+


For a 1.27 M solution of Na2SO4 with a volume of 343 mL, there are 0.872 moles of Na+ ions present.

To summarize:

  1. Write the balanced equation for sodium sulfate dissociation
  2. Identify the given molarity and volume values
  3. Convert volume from mL to L
  4. Use the molarity equation to calculate moles of Na2SO4
  5. Relate moles of Na2SO4 to moles of Na+ using the mole ratio

By following these steps methodically, we can reliably calculate the moles of Na+ in any sodium sulfate solution where we know the molarity and volume. Understanding these types of molarity calculations is essential for success in chemistry courses.

Practice Examples

Let’s look at a couple more examples to solidify understanding of this type of problem:

Example 1

Calculate the moles of Na+ in 250. mL of a 0.55 M Na2SO4 solution.


  1. Na2SO4 → 2Na+ + SO42-
  2. M = 0.55 M
    V = 250 mL = 0.250 L
  3. Moles Na2SO4 = M x L
    = 0.55 mol/L x 0.250 L
    = 0.138 mol Na2SO4
  4. Moles Na+ = 0.138 mol Na2SO4 x (2 mol Na+ / 1 mol Na2SO4)
    = 0.276 mol Na+

Example 2

What volume of a 0.825 M Na2SO4 solution contains 0.651 moles of Na+?


  1. Na2SO4 → 2Na+ + SO42-
  2. M = 0.825 M
    Moles Na+ = 0.651 mol
  3. Moles Na2SO4 = Moles Na+ x (1 mol Na2SO4 / 2 mol Na+)
    = 0.651 mol Na+ x (1 mol Na2SO4 / 2 mol Na+)
    = 0.326 mol Na2SO4
  4. M = moles / L
    0.825 mol/L = 0.326 mol Na2SO4 / L
    L = 0.326 mol Na2SO4 / 0.825 mol/L
    = 0.395 L = 395 mL

So for 0.651 moles Na+, the required volume is 395 mL of a 0.825 M Na2SO4 solution.

Common Mistakes

Some common mistakes students make when solving these types of molarity problems include:

  • Forgetting to convert volume from mL to L
  • Not using the balanced chemical equation to relate moles of solute to moles of ions
  • Mixing up molarity vs. molality units
  • Flipping the mole ratio when relating different moles
  • Using the incorrect volume in the final calculation

Being mindful to avoid these mistakes and double checking your work can help improve success on molarity calculations.

Related Questions

Some other common molarity questions you may encounter are:

  • How do you calculate molarity from moles?
  • How do you find the molarity of a solution?
  • What is the molarity of a solution prepared by dissolving 23.5 grams of sodium chloride in enough water to make 285 mL of solution?
  • How many grams of potassium nitrate are needed to make 250 mL of a 0.100 molar solution?
  • What is the concentration of chloride ions in a 3.26 M solution of iron (III) chloride?

Practice with a variety of molarity calculation types is the key to mastering these important chemistry skills.

Related Formulas

Here are some key formulas to know when working molarity problems:

  • Molarity (M) = moles solute / liters solution
  • Moles solute = Molarity x Volume in liters
  • Volume = Moles solute / Molarity
  • Dilution: M1 x V1 = M2 x V2 (for diluting a solution)


  • M = molarity (mol/L)
  • moles solute = moles of dissolved substance
  • V = volume in liters
  • M1, V1 = initial molarity and volume
  • M2, V2 = final molarity and volume

Keep these formulas handy when solving various molarity problems!

Tips for Solving Molarity Problems

Here are some helpful tips for approaching molarity calculations:

  • Read the problem carefully to identify the known values and determine what the question is asking you to find.
  • Write out the balanced chemical equation for the reaction or dissolution occurring.
  • Convert all volumes into liters and make sure all units are consistent.
  • Identify the correct formula to use based on what the question provides and what it is asking.
  • Plug the known values into the formula and solve carefully showing all units.
  • Check that your answer makes sense logically and that the units are correct.

With practice, a systematic approach, and paying close attention to units, molarity calculations don’t have to be intimidating!

Common Ion Effect and Buffers

Two advanced applications of molarity worth mentioning are the common ion effect and buffers.

The common ion effect refers to the shift in solubility equilibrium when a soluble compound containing an ion already present in solution is added. For example, adding sodium chloride to a solution already containing dissolved sodium ions will decrease the solubility of other sodium salts. Understanding the common ion effect helps explain solubility behaviors.

Buffers contain a weak acid and its conjugate base, which helps resist changes in pH when small amounts of acid or base are added. The concentrations of the weak acid (HA) and conjugate base (A-) are related to the acid dissociation constant (Ka) for the weak acid. The Henderson-Hasselbalch equation is used to calculate pH of a buffer solution based on the molarities and pKa.

While molarity calculations provide the foundation, appreciating these more advanced applications requires a deeper understanding of equilibrium chemistry.

Practice Problems

Here are some additional practice problems with solutions to test your understanding:

Problem 1

What is the molarity of a 2.44 M solution of sodium acetate that was diluted by adding 145 mL of the stock solution to 315 mL of water?


  1. M1V1 = M2V2
    2.44 M x V1 = M2 x (145 mL + 315 mL)
  2. V1 = Initial volume of stock solution, unknown
  3. M2 = Final molarity after dilution, unknown
  4. V2 = Final volume after dilution = 145 mL + 315 mL = 460 mL
  5. 2.44 M x V1 = M2 x 0.460 L
  6. V1 = (M2 x 0.460 L) / 2.44 M
  7. Let V1 = 100 mL
    Then: M2 = (2.44 M x 0.100 L) / 0.460 L = 1.06 M

The final molarity after dilution is 1.06 M.

Problem 2

What volume of a 0.325 M barium nitrate solution contains 3.63 grams of barium nitrate?


  1. Ba(NO3)2 → Ba2+ + 2NO3-
  2. MW of Ba(NO3)2 = 261.34 g/mol
  3. Moles Ba(NO3)2 = 3.63 g / 261.34 g/mol = 0.0139 mol
  4. M = 0.325 mol/L
  5. M = moles / L
    0.325 mol/L = 0.0139 mol / L
    L = 0.0139 mol / 0.325 mol/L
    = 0.0428 L = 42.8 mL

The volume needed is 42.8 mL of the 0.325 M barium nitrate solution.


In summary, calculating molarities from solution concentrations is an essential chemistry skill. By understanding how to interconvert between molarity, volume, and moles of solute, many stoichiometry problems can be tackled systematically. Learning to pay close attention to units and set up problems methodically takes practice but allows you to reliably solve a variety of molarity calculations. Appreciating more advanced applications to solubility and buffers provides further insight into the importance of molar concentrations in chemistry.

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