Question

1A. Determine the pH of a KOH solution made by mixing 0.251 g KOH (s) with enough water to make 1.00 × 102 mL of solution. Assume 100% purity for the KOH (s).

1B. Predict the direction in which the equilibrium will lie for the following reaction:     H3PO4(aq) + HSO4−(aq) H2PO4−(aq) + H2SO4(aq).    Ka1(H3PO4) = 7.5 × 10−3;

Ka(H2SO4) = very large    

2. For H3PO4, Ka1 = 7.3 × 10−3, Ka2 = 6.2 × 10−6, and Ka3 = 4.8 × 10−13. An aqueous solution of Na3PO4 therefore would be?

2B. Calculate the pH of a solution of 0.10 M HA- (aq). The acid H2A has Ka1 = 6.5 × 10−2 and Ka2 = 6.1 × 10–5.  

Answer 1

1A. normality of KOH = w/mwt*1000/V

                     = (0.251/56.1)*(1000/100)

                      = 0.45 N

      pH = 14- (-log(OH-))

         = 14- (-log0.45)

         = 13.65

1B. as H2SO4 is strong acid , it can donates proton easily , so that equilibrium shifts leftside

2. Na3PO4 - Is a salt of weakacid,strongbase . so that it is basic in nature.

2B. the pH of HA- depends upon ka2 only

   so that,

   pH = 1/2(pka2-logC)

   pka2 = -logka2 = -log(6.1*10^-5) = 4.215

      = 1/2(4.215 - log0.1)

      pH = 2.6075