EXPERIMENT 16 ACIDS AND BASES: COMPLETE CURVE ANALYSIS HPOs and .RMNaOH For this

Question

EXPERIMENT 16 ACIDS AND BASES: COMPLETE CURVE ANALYSIS HPOs and .RMNaOH For this table, read the starting pH from the data table and the values at the EPs and SEPs from the graph Initial Equivalence Points Equivalence Points mL pH [H1 mL pH pH 14.bmt | H.6 || 2.3 2.6 2.1s .00 6) Calculate the molarity of the phosphoric acid solution. Calculate Kal from the 1st ½ equivalence point. Calculate Kai from the ionization constant expression Calculate K from the 2hd equivalence point. Calculate the mL of NaOH to the 3d EP using the volume and molarity of the acid Calculate the pH at the 3d EP using an ICE table and the ionizati on constant expression (Ka = 4.8 x 10-13) Check: Confirm the pH at l"and 2nd ES using the pK, values from the ½EPs (K43-48x10-13) pH at 2d EP pH at 1 EP

Explanation / Answer

10.6 mL of 0.842 M NaOh is required to neutralize the acid.

moles of naOH used = 0.842 * 10.6 /1000 = 8.92 * 10^-3

NaOH : H3PO4 ration is 3:1 based on the following reaction

3 NaOH + H3PO4 --------> Na3PO4 + H2O

moles of H3PO4 present = 2.97 * 10^-3

molarity of the acid solution = 2.97 * 10^-3 * 1000/ volume of acid taken

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at first half neutralization point pH = pKa = 2.6

Ka = 2.51 * 10^-3

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at second half neutralization point pH = Pka2

Ka2 = 7.94 *10^-8

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volume of acid used is not mentioned in the problem. if it is known, mL of NaOH can be determined as follows

M (acid) *V (acid) = M (base) * V (base)

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