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Marat540 [252]
3 years ago
12

How many milliliters of 0.10 m pb(no3)2 are required to react with 75 ml of 0.20 m nai? pb(no3)2 + 2 nai --> pbi2 + 2 nano3?

Chemistry
1 answer:
Tom [10]3 years ago
7 0

The Ml of 0.10 Pb(NO3)2 that are required to react with 75 Ml of 0.20M Nai is 75 ml



calculation

Pb(NO3)2 +2Nai → Pbi2 + 2NaNo3


find the number of moles of Nai used

moles= molarity x volume in liters

volume in liters = 75/1000=0.075 moles


mole =0.20 x0.075 = 0.015 moles


by use of mole ratio between Pb(NO3)2 to Nai which is 1 :2 the moles of Pb(NO3)2 = 0.015 x1/2 =7.5 x10^-3 moles


volume (ml) )= moles/molarity x 1000


=volume = (7.5 x10^-3) / 0.10 = 75 Ml

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Answer:

Explanation:

Given parameters:

           pH = 3.50

Unknown:

    concentration of [H₃0⁺] = ?

    concentration of [OH⁻] = ?

Solution:

In order to find the unknown, we use some simple expressions which best explains the pH scale and the equilibrium systems of aqueous solutions.

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         [H₃O⁺] = inverse log₁₀ (-pH) = 10^{-pH} = 10^{-3.5}

          [H₃O⁺] = 3.2 x 10⁻⁴moldm⁻³

       

For the  [OH⁻]:

       we use : pOH = -log₁₀ [OH⁻]

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                  pOH = 14 - pH = 14 - 3.5 = 10.5

  Now we plug the value of pOH into pOH = -log₁₀ [OH⁻]

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The solution is acidic as the concentration of H₃0⁺ is more than that of the OH⁻ ions.

                   

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