Unknown Equlibrium Constant: Calculate the equilibrium constant for the reaction

Question

Unknown Equlibrium Constant:

Calculate the equilibrium constant for the reaction H2PO4 – (aq) + 2 CO3 2– (aq) PO4 3– (aq) + 2 HCO3 – (aq)

Acid Dissociation Constants in Water at 25°C Ka/M Acid Benzoic Acid Carbonic acid Formula C,H6o2 6.3 x 105 H2CO3 4.5 x 10 4.7 x 10-11 6.2 x 10-10 6.3 x 10-4 5.1 x 10-12 4.20 6.35 10.33 9.21 3.20 Hydrocyanic acid HCN Hydrofluoric acid HF Hypochlorous acid HOCI Lactic acid Nitrous acid Phosphoric acid H3PO4 HC3H,O3 1.4 x 10 HNO2 3.85 3.25 2.16 7.21 -4 5.6 x 10 6.9 x 10 6.2 x 108 4.8 x 10-13 Base Protonation Constants in Water at 25°C Kb/MpKb Base Ammonia Aniline Bupivacaine Methylamine Formula NH3 C6H5 NH2 7.4 x 10-10 C18H28N20 1.5 x 10-6 CH3NH2 4.6 x 104 1.8 x 105 4.74 9.13 5.83 3.34

Explanation / Answer

two reactions that combines are---

1.) H2PO4-(aq) --> PO43-(aq) + 2H+ and 2.) 2CO32-(aq) + 2H+ --> 2HCO3-(aq

For 1.)it comes with these two eqaution H2PO4-(aq) --> HPO42- + H+ and HPO42- --> PO43- + H+

so, Ka = product of these two equations Ka i.e,

Ka = 6.2 x 10^-8 x 4.8 x 10^-13 = 2.976 x 10^-20

For 2,) it is the reverse reaction with multiplying factor 2 so,

Ka' = (1/Ka)^2 = (1/4.7 x 10^-11)^2 = 4.527 x 10^20

combining these two equation Ka = Ka x Ka' = 4.527 x 10^20 x 2.976 x 10^-20

equilibrium constant for the reaction Ka = 0.1347

if you satisfied please like it.. thanks..

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