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Let an n-protic acid be an acid that can donate n hydrogen ions and has the formula HnX. If 0.600 L of 0.400 M sodium hydroxide is required to titrate 0.400 L of a 0.300 M HnX to the equivalence point, what is n in HnX?

Respuesta :

superman1987

Answer:

2, the acid is H₂X.

Explanation:

  • It is known at equivalence point: the no. of millimoles of base is equal to the no. of millimoles of acid.

∴ (nMV) of NaOH = (nMV) for HnX.

where, n is the no. of producible H⁺ or OH⁻ of the acid or base, respectively.

M is the molarity of the acid or base.

V is the volume of the acid or base.

  • For NaOH:

n = 1, M = 0.4 M, V = 0.6 L.

  • For HnX:

n = ???, M = 0.3 M, V = 0.4 L.

∴ n for HnX = (nMV) of NaOH / (MV) for HnX = (1)(0.4 M)(0.6 L)/(0.3 M)(0.4 L) = 2.

∴ the acid is H₂X.

batolisis

The value of n in the n-protic acid  (HnX)  is ;  2

Given data :

Formula of the n-protic acid is ;  HnX

Volume of  sodium hydroxide ( V ) = 0.600 L

Molarity of base ( M ) = 0.400 M

n for base ( Sodium hydroxide )  = 1

Volume of acid = 0.400 L

molarity of acid = 0.300 M

n for acid = ?

At equivalence point

( nMV) of NaOH = (nMV) for HnX. ----- ( 1 )

n = number of ions

M = molarity

V = volume

Next step : make n for acid subject of the formula above

n =  ( nMV ) / ( MV ) for acid

   = ( 1 * 0.400 * 0.600 ) / ( 0.300 * 0.400 )

∴ n = 2

Hence we can conclude that the value of n in HnX is = 2.

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