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If 75.0 mL of a 0.20 M solution of sodium nitrate (NaNO3) is mixed with 25.0 mL of 0.10 M barium nitrate (Ba(NO3)2), what is the molar concentration of nitrate in the resulting solution? A. 0.10 M B. 0.15 M C. 0.20 M D. 0.18 M E. 0.30 M

Respuesta :

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Answer:

C. 0.20 M

Explanation:

To solve this problem we calculate the moles of nitrate (NO₃⁻) that both of those solutions contain:

  • Nitrate from NaNO₃:

75.0 mL * 0.20 M = 15 mmol NaNO₃ = 15 mmol NO₃⁻

  • Nitrate from Ba(NO₃)₂:

25.0 mL * 0.10 M = 2.5 mmol Ba(NO₃)₂

2.5 mmol Ba(NO₃)₂ * [tex]\frac{2mmolNO_{3}}{1mmolBa(NO_{3})_{2}}[/tex] = 5.0 mmol NO₃⁻

  • So the final number of NO₃⁻ moles is (15 + 5) 20 mmol.
  • The final volume is (75 + 25) 100 mL.

So the molar concentration is:

  • 20 mmol / 100 mL = 0.20 M

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