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Calculate The Ph For Each Of The Following Cases

in the titration of 50.0 mL of 0.180 M HClO(aq) with 0.180 M KOH(aq). The ionization constant for HClO can be found here. (a) before addition of any KOH (b) after addition of 25.0 mL of KOH (c) after addition of 30.0 mL of KOH (d) after addition of 50.0 mL of KOH (e) after addition of 60.0 mL of KOH
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pH is determined by the dissociation of HCIO HCIO -> H+ CIO- I .18 0 0 C -x +x +x E .18-x x x Ka=[CIO-][H+]/[HCIO]= x^2/.18M-x=x^2/.18 x=[H+]=.000085 pH=4.07 with equal concentrations of monoprotic titrant and analyte, the EQ point would occur when the volumes are equal. 50mL KOH at 25mL, this is the first midpoint. thus, pH=pKa=-log(4e-8)=7.4 At 30 mL we are at 30/50 concentration so 60 percent. this means that 40 percent of the acid still remains and 60 percent of the conjugate base has formed. thus the pH=7.4+log(60/40)=7.57 At 50 mL, we are at the EQ point. all of the acid has reacted. thus the pH is determined by the base. 50mL*.18M=9mmol CIO- [CIO-]=9mmol/100mL=.09 M Kb=1e-14/4e-8=2.5e-7 [OH-]=sqrt(.09*2.5e-7)=3.82 pH=10.18 At 60 mL is beyond EQ, excess KOH. we can ignore CIO- because KOH is a strong base. 50 mL reacted with the acid, so only 10mL is used in the calculations. m1v1=m2v2 m2=.18M*10ml/60ml+50ml .0164 M OH- pOH=1.786 pH=12.21
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