Question

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2. A 20.00 mL sample of H3PO4 (aq) requires 56,01 mL of 0.1885 M NaOH for titration to the third equivalence point. a) What is the molarity of the H3PO. (aq) sample? (4 pts) b) What is the trend in K, values going from H3PO, to H,POto HPO}? What is the reason behind this trend? (2 pts)

Answer 1

(9) At equivalence point H₃ PO 4 an+ 3 *99 an) Nao H - Naz POq + 3 H2Oron 297 = 3 * No. of millimoles.of Naoh .No.of milli moles of S H3 PO 4 e Mava := 3 Mb Vb Ma = 3 Mb Vo Va where. Ma = Molarity of acid (H3PO4) Mb = Molasity of base (NaOH) = 0.1885 m Va - Volume of acid - 20.00 ML Vn - Volume of base = 56.0 ml Ma = 1.58 M = 3x0, 1885 M x 56.01 mL 20 ml Therefore, the molarity of H3PO4 is approximately 1.58 m.

(b) H3PO4 is a triprotic acid which can undergo three dissociations and will therefore have three dissociation constants in the following thend- 1 H3PO A (99) H2POA GA HROA GA kap > keq,) Haz Hrani + H2PO a can) kap - Hags + APO on kaz Htap) 4 80 9 169) . Kaz 'This trend te keap) kaz) kas is always true for this or any polyprotic acids ie each ionization step is more difficult because it is more difficult to remove Ht from a molecule as its negative change increases.