Asked by Cicily
                Question as follows:
75.0g tartaric acid in 1 Liter is what molarity(moles/liter) mw=150.087 g/mol
M1=
25 ml tartaric solution is titrated with 31 ml ammonium.
What is the equation rearanged for normality of ammonium: starting with N1V1=N2V2
N1=
V1=
V2=
Final Answer N2=
THANKS! I can't wrap my head around this!
            
        75.0g tartaric acid in 1 Liter is what molarity(moles/liter) mw=150.087 g/mol
M1=
25 ml tartaric solution is titrated with 31 ml ammonium.
What is the equation rearanged for normality of ammonium: starting with N1V1=N2V2
N1=
V1=
V2=
Final Answer N2=
THANKS! I can't wrap my head around this!
Answers
                    Answered by
            DrBob222
            
    M = mols/L and mols = g/molar mass.
Therefore, mols H2T(tartaric acid) = 75.0/150.087 = ?
and M = mols/L.
Nomality of H2T = 2*M
Then 25*N H2T = 31*N NH3. Solve for N NH3.
By the way I assume this is ammonia, the ammonium ion doesn't exist by itself.
    
Therefore, mols H2T(tartaric acid) = 75.0/150.087 = ?
and M = mols/L.
Nomality of H2T = 2*M
Then 25*N H2T = 31*N NH3. Solve for N NH3.
By the way I assume this is ammonia, the ammonium ion doesn't exist by itself.
                    Answered by
            Cicily
            
    Thank you DrBob222!
    
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