Ol ionic 3Ba 2 aq 6OH aq 6H aq 2PO 4 3 aq Ba 3 PO 4 2 s 6H 2 Ol net ionic 3Ba 2

Ol ionic 3ba 2 aq 6oh aq 6h aq 2po 4 3 aq ba 3 po 4 2

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O(l) ionic:3Ba 2+(aq) + 6OH-(aq) + 6H-(aq) + 2PO43-(aq) Ba3(PO4)2(s) + 6H2O(l)net ionic:3Ba 2+(aq) + 2PO43-(aq) Ba3(PO4)2(s)c.HClO4+ Mg(OH)2balanced:2HClO4(aq) + Mg(OH)2(s) Mg(ClO4)2(aq) + 2H2O(l) ionic:2H+(aq) + 2ClO4-(aq) + Mg2(aq)+2OH(aq) Mg2+(aq) + 2ClO4-(aq) + 2H2O(l)net ionic:2H+(aq)+2OH-(aq) 2H2O(l)5.Identify the acid-base conjugate pairs in each of the following reactions: (5 points)a.CH3COO+ HCN CH3COOH + CNBase 1Acid1Acid2Base2b.HCO3+ HCO3H2CO3+ CO3–2c.H2PO4+ NH3HPO42–+ NH4+Acid1Base1AcidAcid2d.HClO + CH3NH2CH3NH3++ ClOAcid1Base1Acid2Base2e.CO32–+ H2O HCO3+ OHBase1Acid1Acid2Base2Acid1Base1Acid2Base2(Reference: Chang 15.5)6.Calculate the pH of each of the following solutions: (2 points)2Copyright © 2014 by Thomas Edison State College. All rights reserved.
[H+]=0.0010 M = 1 x 10-3p H = -log [1 x 10-3] = 3pH of HCL = 3KOH(aq) K+(aq) + OH-(aq)[OH-] = 0.76 M = 76 x 10-2p OH = -log [76 x 10-2] = 2-log 76 = 2-1.88 = 0.119pH = 14-0.119 = 13.881pH=13.88(Reference: Chang 15.17)7.Calculate the hydrogen ion concentration in mol/L for solutions with the following pH values: (4 points)pH = 2.42pH = -log [H+]2.42 = -log [H+][H+] = antilog of -2.42 = 3.80 x 10-3 mol/LpH = 11.21pH = -log [H+]11.21 = -log [H+][H+] = antilog of -11.21 = 6.16 x 10-12mol/LpH = 6.96pH = -log [H+]6.96 = -log [H+][H+] = antilog of -6.96 = 1.09 x 10-7mol/LpH = 15.00pH = -log [H+]15.00 = -log [H+][H+] = antilog of -15.00 = 1 x 1015 mol/L(Reference: Chang 15.19)3Copyright © 2014 by Thomas Edison State College. All rights reserved.

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