You have a 50.0mL solution of 1.35M NH_3 which you are titrating with a 0.75M so

Question

You have a 50.0mL solution of 1.35M NH_3 which you are titrating with a 0.75M solution of HBr (K_B(nh3)= 1.8 times 10^-5) What is the pH before you start titration? Suppose that 0.1000 mole each of h_2 and i_2 are placed in 1.000-L flask, stoppered, and the mixture is heated to 425degree C. At equilibrium, the concentration of I_2 is found to be 0.0210 M. What are the equilibrium concentrations of and HI, respectively? Calculate K_c for the following reaction at 425degree C. H_2(g) + i_2(g) 2 HI(g)

Explanation / Answer

7.

no of moles of NH3 = molarity * volume in L

                            = 1.35*0.05   = 0.0675moles

no of moles of HBr = molarity * volume in L

                             = 0.75*0.05 = 0.0375 moles

         NH3     + HBr ------> NH4Br

I      0.0675          0.0375     0

C   -0.0375        -0.0375       0.0375

E    0.03             0              0.0375

     Pkb = -logkb

            = -log1.8*10-5

           = 4.74

POH   = PKb + log[NH4Br]/[NH3]

          = 4.74 + log0.0375/0.03

          = 4.74 + 0.0969   = 4.8369

PH    = 14-POH

        = 14-4.8369 = 9.1631 >>>> answer

8.      molarity = no of moles/volume inL

                      = 0.1/1 = 0.1M

[H2] = [I2]   = 0.1M

     H2      + I2 --------> 2HI

I     0.1    0.1         0

C -0.079      -0.079        2*0.079

E 0.021        0.021        0.158

   Kc = [HI]2/[H2][I2]

         = (0.158)2/0.021*0.021

        = 0.024964/0.000441   = 56.6 >>>> answer

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