Calculate the amount of 0.1 M acid needed to neutralize 10,000 liters of pH 8.0 water. The acid is Carbonic Acid (H₂CO₃) please show w

Question

Calculate the amount of 0.1 M acid needed to neutralize 10,000 liters of pH 8.0 water.
The acid is Carbonic Acid (H₂CO₃) please show work!

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Kim Cúc 4 years 2021-08-12T10:14:02+00:00 1 Answers 43 views 0

Answers ( )

  1. danthu
    0
    2021-08-12T10:15:23+00:00
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    Answer:

    50mL of the 0.1M acid are needed to neutralize the solution

    Explanation:

    To solve this question we have to find the moles of OH- present in the basic solution. Then, using the chemical equation:

    2OH- + H2CO3 → 2H2O + CO3²⁻

    We can find the moles of carbonic acid and its volume using the 0.1M solution as follows:

    Moles OH-:

    pH = -log [H+]  

    10^-pH = [H+]  

    [H+] = 1×10⁻⁸M  

    As:  

    [OH-] = Kw / [H+]  

    [OH-] = 1×10⁻¹⁴ / 1×10⁻⁸

    [OH⁻] = 1×10⁻⁶M

    The moles in 10000L are:

    10000L * (1×10⁻⁶moles OH- / L) = 0.01 moles OH-

    Moles H₂CO₃:

    0.01 moles OH- * (1mol H₂CO₃ / 2mol OH⁻) = 0.005 moles H₂CO₃  

    Volume:  

    0.005 moles H₂CO₃ * (1L / 0.1moles) = 0.05L =  

    50mL of the 0.1M acid are needed to neutralize the solution

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