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Thread: molarity problem

  1. #1 molarity problem 
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    1.26g of a hydrated dibasic acid (H2A . nH2O) of anhydrous molecular mass 90.0 was made up to 250 cm^3. 25.0 cm^3 of this solution required 20.0 cm^3 of 0.100M potassium hydroxide solution for neutralization. Calculate the number of molecules of water of crystallization per molecule of the hydrated acid.

    Really need some help on this.


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  3. #2  
    Forum Freshman César's Avatar
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    1) Write down the rection between H2A and potassium hydroxide.
    2) Calculate number of moles of potassium hydroxide in the 20mL of 0.01M solution
    3) According to 1) and 2) calculate number of moles present of H2A in 25mL.
    4) Using 3) calculate number of moles of H2A in 250 mL.
    5) Using 4) and the molecular weight of anhydro H2A calculate grams of H2A.
    6) The difference in weight between the hydrated H2A (1.26g) and the anhydro H2A is the weight of coordination water.
    7) Using 6) calculate number of moles of water
    => Using 7) and 4) calculate number of moles of water per mol of H2A. This result is the answer.



    Best regards,

    César


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  4. #3  
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    H2A . nH20 + 2KOH = H2O + K2A
    no. of moles of KOH= 0.1*0.2 =0.02
    mole ratio, from equation, H2A . nH2O to KOH = 1:2
    H2A... has 0.01 no. of moles.

    1.26/(90+18n)=0.01
    n=2

    I think this is the answer.
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  5. #4  
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    Thanks a lot by the way. There is another molarity question which i want to ask, so i wil post it here as well. This will probably be the last question.
    This is not my homework or anything, its just that im studying by myself as this is not in my school curriculum.

    The formula of a solid dibasic acid is H2X. 2.88g of the acid is dissolved in some distilled water and the solution is then diluted to 250.0 cm^3 with distilled water. 25.0 cm ^3 of the diluted solution requires 16.0 cm^3 of 0.40M sodium hydroxide solution for complete neutralization. What is the molar mass of H2X?
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  6. #5  
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    Go from 1) to 4). Then do the reciprocal of 5)

    Best regards,

    César
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  7. #6  
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    Ok I got up to step 3 and found that H2X has 0.0034 mol in 25cm^3. But how do I find the no. of moles in 250cm^3? Im sure it is not multiplying it by 10.
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  8. #7  
    Forum Freshman César's Avatar
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    Yes, it is.
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  9. #8  
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    ok heres my answer.
    NaOH no. of moles = 0.0064mol
    mole ratio, H2X:NaOH, 1:2
    H2X no. of moles in 25cm^3 = 0.0032mol
    H2X no. of moles in 250cm^3 = 0.032mol
    2.88g/molar mass of H2X = 0.032
    90=molar mass of H2X

    Thanks a lot.
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