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What is the concentration (m) of sodium ions in 4.57 l of a 2.35 m na3p solution?

User Ashareef
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2 Answers

1 vote
Answer: 1.03 x 10^25 ions Na
User Braden Holt
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3 votes

Answer:

7.05 M

Step-by-step explanation:

  • An equation can be written:

Na₃P → 3Na⁺ + P⁻³

  • The formula for concentration (M) is the number of moles divided by the volume in L:

C = n / V

With the concentration and volume given by the problem, we know the number of moles of Na₃P, then with that number we can calculate the moles of Na⁺:

4.57 L * 2.35 M = 10.7395 mol Na₃P

10.7395 mol Na₃P *
(3molNa^(+))/(1molNa_(3)P) = 32.2185 mol Na⁺.

Concentration of Na⁺ = mol Na⁺ / V

[Na⁺] = 32.2185 mol Na⁺ / 4.57 L = 7.05 M

User Derek Lee
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